DISJOINT 5-CYCLES IN A GRAPH

Discussiones Mathematicae Graph Theory 32 (2012) 221–242 doi:10.7151/dmgt.1605

DISJOINT 5-CYCLES IN A GRAPH

Hong Wang Department of Mathematics The University of Idaho Moscow, Idaho, 83844 USA e-mail: [email protected]

Abstract We prove that if G is a graph of order 5k and the minimum degree of G is at least 3k then G contains k disjoint cycles of length 5. Keywords: 5-cycles, pentagons, cycles, cycle coverings. 2010 Mathematics Subject Classification: 05C38, 05C70, 05C75.

1.

Introduction and Notation

A set of graphs is said to be disjoint if no two of them have any common vertex. Corr´ adi and Hajnal [3] investigated the maximum number of disjoint cycles in a graph. They proved that if G is a graph of order at least 3k with minimum degree at least 2k, then G contains k disjoint cycles. In particular, when the order of G is exactly 3k, then G contains k disjoint triangles. Erd˝os and Faudree [5] conjectured that if G is a graph of order 4k with minimum degree at least 2k, then G contains k disjoint cycles of length 4. This conjecture has been confirmed by Wang [8]. El-Zahar [4] conjectured that if G is a graph of order n = n1 + n2 + · · · + nk with ni ≥ 3 (1 ≤ i ≤ k) and the minimum degree of G is at least ⌈n1 /2⌉ + ⌈n2 /2⌉ + · · · + ⌈nk /2⌉, then G contains k disjoint cycles of lengths n1 , n2 , . . . , nk , respectively. He proved this conjecture for k = 2. When n1 = n2 = · · · = nk = 3, this conjecture holds by Corr´ adi and Hajnal’s result. When n1 = n2 = · · · = nk = 4, El-Zahar’s conjecture reduces to the above conjecture of Erd˝os and Faudree. Abbasi [1] announced a solution to El-Zahar’s conjecture for very large n. In this paper, we develop a constructive method to show that El-Zahar’s conjecture is true for all n = 5k with ni = 5 (1 ≤ i ≤ k).

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Theorem 1. If G is a graph of order 5k and the minimum degree of G is at least 3k, then G contains k disjoint cycles of length 5. We shall use the terminology and notation from [2] except as indicated. Let G be a graph. Let u ∈ V (G). The neighborhood of u in G is denoted by N (u). Let H be a subgraph of G or a subset of V (G) or a sequence of distinct vertices of G. We define N (u, H) to be the set of neighbors of u contained in H, and let e(u, H) = |N (u, H)|. Clearly, N (u, G) = N (u) and e(u, G) is the degree of u in G. If X is a subgraph of G or a subset of V (G) or a sequence of distinct P vertices of G, we define N (X, H) = ∪u N (u, H) and e(X, H) = u e(u, H) where u runs over all the vertices in X. Let x and y be two distinct vertices. We define I(xy, H) to be N (x, H) ∩ N (y, H) and let i(xy, H) = |I(xy, H)|. Let each of X1 , X2 , . . . , Xr be a subgraph of G or a subset of V (G). We use [X1 , X2 , . . . , Xr ] to denote the subgraph of G induced by the set of all the vertices that belong to at least one of X1 , X2 , . . . , Xr . We use Ci to denote a cycle of length i for all integers i ≥ 3, and use Pj to denote a path of order j for all integers j ≥ 1. For a cycle C of G, a chord of C is an edge of G − E(C) which joins two vertices of C, and we use τ (C) to denote the number of chords of C in G. Furthermore, if x ∈ V (C), we use τ (x, C) to denote the number of chords of C that are incident with x. For each integer k ≥ 3, a k-cycle is a cycle of length k. If S is a set of subgraphs of G, we write G ⊇ S. For an integer k ≥ 1 and a graph G′ , we use kG′ to denote a set of k disjoint graphs isomorphic to G′ . If G1 , . . . , Gr are r graphs and k1 , . . . , kr are r positive integers, we use k1 G1 ⊎ · · · ⊎ kr Gr to denote a set of k1 + · · · + kr disjoint graphs which consist of k1 copies of G1 , . . . , kr−1 copies of Gr−1 and kr copies of Gr . For two graphs H1 and H2 , the union of H1 and H2 is still denoted by H1 ∪ H2 as usual, that is, H1 ∪ H2 = (V (H1 ) ∪ V (H2 ), E(H1 ) ∪ E(H2 )). Let each of Y and Z be a subgraph of G, or a subset of V (G), or a sequence of distinct vertices of G. If Y and Z do not have any common vertices, we define E(Y, Z) to be the set of all the edges of G between Y and Z. Clearly, e(Y, Z) = |E(Y, Z)|. If C = x1 x2 . . . xr x1 is a cycle, then the operations on the subscripts of the xi ’s will be taken by modulo r in {1, 2, . . . , r}. We use B to denote a graph of order 5 and size 6 such that B has two edgedisjoint triangles. We use F to denote a graph of order 5 and size 5 such that F has a vertex of degree 1 and a 4-cycle. Let F1 be the graph of order 5 obtained from F by adding a new edge to F such that the new edge joins the two vertices of F whose degrees in F are 2. Let F2 be the graph of order 5 and size 7 obtained from K2,3 by adding a new edge to K2,3 such that F2 has two adjacent vertices of degree 4. We use K4+ to denote the graph of order 5 and size 7 such that K4+ has a vertex of degree 1. Finally, we use K5− to denote a graph of order 5 with 9 edges. Let {H, L1 , . . . , Lt } be a set of t+1 disjoint subgraphs of G such that Li ∼ = C5

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for i = 1, . . . , t. We say that {H, L1 , . . . , Lt } is optimal if for any t + 1 disjoint [H, L1 , . . . , Lt ] with H ′ ∼ subgraphs H ′P , L′1 , . . . , L′t inP = C5 (1 ≤ i ≤ t), = H and L′i ∼ t t ′ we have that i=1 τ (Li ) ≤ i=1 τ (Li ). Let L be a 5-cycle of G and H a subgraph of order 5 in G. We write H ≥ L if H has a 5-cycle L′ such that τ (L′ ) ≥ τ (L). Moreover, if τ (L′ ) > τ (L), we write H > L. Let L be a 5-cycle of G. Let u ∈ V (L) and x0 ∈ V (G) − V (L). We write x0 → (L, u) if [L − u + x0 ] ⊇ C5 . Moreover, if [L − u + x0 ] ≥ L then we a write x0 ⇒ (L, u) and if [L − u + x0 ] > L then we write x0 → (L, u). In a na addition, if it does not hold that x0 → (L, u) then we write x0 → (L, u). Clearly, a x0 ⇒ (L, u) when x0 → (L, u). If x0 → (L, u) for all u ∈ V (L) then we write a x0 → L. Similarly, we define x0 ⇒ L and x0 → L. If [L − u + x0 ] ⊇ B, we write z x0 → (L, u). Let P be a path of order at least 2 or a sequence of at least two distinct vertices in G − V (L + x0 ). Let X be a subset of V (G) − V (L + x0 ) with |X| ≥ 2. We write x0 → (L, u; P ) if x0 → (L, u) and u is adjacent to the two end vertices of P . In this case, if P is a path of order 4, then [x0 , L, P ] ⊇ 2C5 . We write x0 → (L, u; X) if x0 → (L, u; xy) for some {x, y} ⊆ X with x 6= y. We write x0 → (L; P ) if x0 → (L, u; P ) for some u ∈ V (L) and x0 → (L; X) if x0 → (L, u; X) for z z some u ∈ V (L). Similarly, we define the notation x0 → (L; P ) and x0 → (L; X). z nz If it does not hold that x0 → (L; P ), we write x0 → (L; P ). If it does not hold z nz that x0 → (L; X), we write x0 → (L; X). 2. 2.1.

Sketch of the Proof of Theorem 1 and Preliminary Lemmas Sketch of the proof of Theorem 1

Let G be a graph of order 5k with minimum degree at least 3k. Suppose, by way of contradiction, that G 6⊇ kC5 . We may assume that G is maximal, i.e., G+xy ⊇ kC5 for each pair of non-adjacent vertices x and y of G. Thus G ⊇ P5 ⊎ (k − 1)C5 . Our first goal is to show that G ⊇ K4+ ⊎ (k − 1)C5 . This will be accomplished through a series of lemmas in Section 2.2. Say G ⊇ {D, L1 , . . . , Lk−1 } with D ∼ = C5 (1 ≤ i ≤ k). Let x0 ∈ V (D) with e(x0 , D) = 1 and let = K4+ and Li ∼ Q = D − x0 . We shall estimate the upper bound on 2e(x0 , G) + e(Q, G) ≥ 18k. This needs an estimation on each 2e(x0 , Li ) + e(Q, Li ). The idea is to show that if e(x0 , Li ) is increasing then e(Q, Li ) is decreasing for otherwise [D, Li ] ⊇ 2C5 , a contradiction. This is accomplished in Lemma 3.3. It turns out that 2e(x0 , G) + e(Q, G) < 18k, a contradiction. 2.2.

Preliminary lemmas

Our proof of Theorem 1 will use the following lemmas. Let G = (V, E) be a given graph in the following.

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Lemma 2.1. The following statements hold: (a) If P ′ and P ′′ are two disjoint paths of G such that |V (P ′ )| = 2, 2 ≤ |V (P ′′ )| ≤ 3 and e(P ′ , P ′′ ) ≥ 3, then [P ′ , P ′′ ] ⊇ C4 . (b) If x and y are two distinct vertices and P is a path of order 3 in G such that {x, y} ∩ V (P ) = ∅ and e(xy, P ) ≥ 5, then [x, y, P ] contains a 5-cycle C such that τ (C) ≥ 2. (c) If D is a graph of order 5 with e(D) ≥ 7, then D ⊇ C5 , unless D ∼ = K4+ or ∼ F2 . D= (d) If R is a subset of V (G) and L is a 5-cycle of G − R such that |R| = 4 and e(R, L) ≥ 13, then u → (L; R − {u}) for some u ∈ R, or there exist two labellings R = {y1 , y2 , y3 , y4 } and L = b1 b2 b3 b4 b5 b1 such that N (y1 , L) = N (y2 , L) = {b1 , b2 , b3 , b4 }, N (y3 , L) = {b1 , b5 , b4 } and N (y4 , L) = {b1 , b4 }. Proof. It is easy to check (a), (b) and (c). To prove (d), we suppose, for a contradiction, that u 6→ (L; R − {u}) for all u ∈ R. Let R = {y1 , y2 , y3 , y4 } be such that e(y1 , L) ≥ e(yi , L) for all yi ∈ R. As e(R, L) ≥ 13, e(y1 , L) ≥ 4 and there exists b ∈ V (L) such that e(b, R − {y1 }) ≥ 2. If e(y1 , L) = 5 then y1 → (L, b; R − {y1 }), a contradiction. Hence we may assume that L = b1 b2 b3 b4 b5 b1 and e(y1 , b1 b2 b3 b4 ) = 4. Thus e(bi , R − {y1 }) ≤ 1 for i ∈ {2, 3, 5}. Then 6 ≥ e(b1 b4 , R − {y1 }) ≥ 13 − 4 − 3 = 6. It follows that e(b1 b4 , R − {y1 }) = 6 and e(bi , R − {y1 }) = 1 for i ∈ {2, 3, 5}. W.l.o.g., say b2 y2 ∈ E. Then e(b3 , y3 y4 ) = 0 as y2 6→ (L, b3 ; R − {y2 }). Hence b3 y2 ∈ E. W.l.o.g., say b5 y3 ∈ E. Thus (d) holds. Lemma 2.2. Let D and L be disjoint subgraphs of G such that D ∼ = B and ∼ L = C5 . Say D = x0 x1 x2 x0 x3 x4 x0 . Suppose that e(D − x0 , L) ≥ 13. Then [D, L] ⊇ 2C5 . Proof. Let H = [D, L]. On the contrary, suppose H 6⊇ 2C5 . Then it is easy to see that xi 6→ (L; xj xs ) and xi 6→ (L; xj xt ) for (1)

{{i, j}, {s, t}} = {{1, 2}, {3, 4}}.

Let R = {x1 , x2 , x3 , x4 }. W.l.o.g., say e(x1 , L) ≥ e(xi , L) for all xi ∈ R. Then e(x1 , L) ≥ 4. First, assume that e(x1 , L) = 5. By (1), I(x2 x3 , L) = I(x2 x4 , L) = ∅. Thus e(x2 x3 , L) ≤ 5 and e(x2 x4 , L) ≤ 5. Since e(R, L) ≥ 13, it follows that e(x4 , L) ≥ 3 and e(x3 , L) ≥ 3. As x3 6→ (L; x1 x4 ), we see that e(x3 , L) = 3. Similarly, e(x4 , L) = 3. Then e(x2 , L) = 2. As x2 6→ (L; x1 x3 ), we see that the two vertices of N (x2 , L) must be consecutive on L. Say N (x2 , L) = {a1 , a2 }. Then [x0 , x1 , x2 , a1 , a2 ] ⊇ C5 and [x3 , x4 , a3 , a4 , a5 ] ⊇ C5 , a contradiction. Therefore e(x1 , L) = 4. Say N (x1 , L) = {a1 , a2 , a3 , a4 }. By (1), I(x2 xj , {a2 , a3 , a5 }) = ∅ for j ∈ {3, 4}. Thus e(x2 xj , L) ≤ 7 for j ∈ {3, 4} and so e(xj , L) ≥ 2 for j ∈ {3, 4}.

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First, assume e(x2 xj , L) = 7 for some j ∈ {3, 4}. Say e(x2 x3 , L) = 7. Then I(x2 x3 , L) = {a1 , a4 } and e(ai , x2 x3 ) = 1 for i ∈ {2, 3, 5}. If e(x4 , a2 a3 ) ≥ 1, say w.l.o.g. x4 a2 ∈ E, then [a1 , a2 , x4 , x0 , x3 ] ⊇ C5 and so x2 a5 6∈ E as H 6⊇ 2C5 . Consequently, x3 a5 ∈ E and so H ⊇ 2C5 = {x3 a5 a1 a2 x4 x3 , x1 x0 x2 a4 a3 x1 }, a contradiction. Hence e(x4 , a2 a3 ) = 0 and so e(x4 , a1 a4 ) ≥ 1. W.l.o.g., say x4 a1 ∈ E. Then [x3 , x4 , a1 , a5 , a4 ] ⊇ C5 and so e(x2 , a2 a3 ) = 0 as H 6⊇ 2C5 . Thus e(x3 , a2 a3 ) = 2. As e(x3 , L) ≤ e(x1 , L) = 4, x3 a5 6∈ E. Thus x2 a5 ∈ E, and consequently, H ⊇ 2C5 = {x3 x4 a1 a2 a3 x3 , x1 x0 x2 a5 a4 x1 }, a contradiction. Therefore e(x2 xj , L) ≤ 6 for j ∈ {3, 4} and so e(xj , L) ≥ 3 for j ∈ {3, 4}. Similarly, if e(x3 , L) = 4 then e(x1 x4 , L) ≤ 6, a contradiction. Hence e(x3 , L) = 3. Similarly, e(x4 , L) = 3. Then e(x2 , L) = 3 as e(R, L) ≥ 13. Assume x2 a5 ∈ E. Then e(a5 , x3 x4 ) = 0 by (1). As e(x3 x4 , L) = 6, either e(x3 x4 , a1 a2 ) ≥ 3 or e(x3 x4 , a3 a4 ) ≥ 3. Say w.l.o.g. the former holds. Then [x3 , x0 , x4 , a1 , a2 ] ⊇ C5 and [x1 , x2 , a5 , a4 , a3 ] ⊇ C5 , a contradiction. Hence x2 a5 6∈ E. As e(x2 , L) = 3, either e(x2 , a1 a3 ) = 2 or e(x2 , a2 a4 ) = 2. W.l.o.g., say the former holds. As x2 6→ (L; x1 xj ) for j ∈ {3, 4}, e(a2 , x3 x4 ) = 0. As e(x3 x4 , L) = 6, either e(x3 x4 , a3 a5 ) ≥ 3 or e(x3 x4 , a1 a4 ) ≥ 3. Thus either [x3 , x4 , a3 , a4 , a5 ] ⊇ C5 or [x3 , x4 , a4 , a5 , a1 ] ⊇ C5 . In each situation, we see that H ⊇ 2C5 , a contradiction. Lemma 2.3. Let P and L be disjoint subgraphs of G such that P ∼ = P5 and L∼ = C5 . Suppose that {P, L} is optimal, e(P, L) ≥ 16 and [P, L] 6⊇ 2C5 . Then [P, L] ⊇ F ⊎ C5 . Proof. Say P = x1 x2 x3 x4 x5 with e(x1 , L) ≥ e(x5 , L) and L = a1 a2 a3 a4 a5 a1 . Then e(x1 , L) ≥ 1. Let H = [P, L]. On the contrary, suppose H 6⊇ F ⊎ C5 . Assume first that e(x1 , L) = 1. Say x1 a1 ∈ E. As e(P, L) ≥ 16 and e(x5 , L) ≤ 1, e (x2 x3 x4 , L) ≥ 14. Thus e (x2 , a3 a4 ) ≥ 1. W.l.o.g., say x2 a3 ∈ E. Then [x1 , x2 , a3 , a2 , a1 ] ⊇ C5 . As e(x3 x4 , L) ≥ 14 − e(x2 , L) ≥ 9, e(x3 x4 , a4 a5 ) ≥ 3. By Lemma 2.1(a), [x5 , x4 , x3 , a4 , a5 ] ⊇ F and so H ⊇ F ⊎ C5 , a contradiction. Hence e(x1 , L) ≥ 2. As e(P, L) ≥ 16, I(x2 x4 , L) 6= ∅ or I(x3 x5 , L) 6= ∅. Therefore x1 6→ L for otherwise H ⊇ F ⊎C5 . Hence e(x1 , L) ≤ 4. We divide the proof into the following cases. Case 1. e(x1 , L) = 4. Say N (x1 , L) = {a1 , a2 , a3 , a4 }. Then [L − ai + x1 ] ⊇ F for all ai ∈ V (L). Thus I(x2 x5 , L) = ∅ as H 6⊇ F ⊎ C5 . As x1 6→ L, τ (a5 , L) = 0. a Then x1 → (L, a5 ). By the optimality of {P, L}, [P − x1 + a5 ] 6⊇ P5 and so e(a5 , x2 x5 ) = 0 and e(a5 , x3 x4 ) ≤ 1. Thus e(x2 x5 , L) ≤ 4 and so e(x3 x4 , L) ≥ 8. Suppose e(x2 , L) ≥ 1. Then e(x2 , a2 a4 ) ≥ 1 or e(x2 , a1 a3 ) ≥ 1. W.l.o.g., say the former holds. Then [x1 , x2 , a2 , a3 , a4 ] ⊇ C5 . As H 6⊇ F ⊎ C5 and by Lemma 2.1(a), we see that e(x3 x4 , a1 a5 ) ≤ 2. It follows that e(x3 x4 , a2 a3 a4 ) = 6 and e(x2 x5 , L − a5 ) = 4. Thus e(a2 , x2 x5 ) > 0. Then [P − x1 + a2 ] ⊇ F . As x1 →

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(L, a2 ), H ⊇ F ⊎C5 , a contradiction. Hence e(x2 , L) = 0. Similarly, if e(x5 , L) = 4 then e(x4 , L) = 0 and so e(P, L) < 16, a contradiction. Hence e(x5 , L) ≤ 3 and so e(x3 x4 , L) ≥ 9. As e(a5 , x3 x4 ) ≤ 1, it follows that e(x3 x4 , L − a5 ) = 8, e(a5 , x3 x4 ) = 1 and e(x5 , L) = 3. Then e(ai , x3 x5 ) = 2 for some i ∈ {2, 3} and so H ⊇ F ⊎ C5 as x1 → (L, ai ), a contradiction. Case 2. e(x1 , L) = 3. Then e(x5 , L) ≤ 3. First, suppose that the three vertices in N (x1 , L) are not consecutive on L. Say N (x1 , L) = {a1 , a2 , a4 }. Clearly, I(x2 x5 , L) ⊆ {a4 } since H 6⊇ 2C5 and H 6⊇ F ⊎ C5 . Hence e(x2 x5 , L) ≤ 6. If x2 a4 ∈ E then [x1 , x2 , a1 , a5 , a4 ] ⊇ C5 . As H 6⊇ F ⊎ C5 , e(x3 x4 , a2 a3 ) ≤ 2. Similarly, [x1 , x2 , a2 , a3 , a4 ] ⊇ C5 and so e(x3 x4 , a1 a5 ) ≤ 2. Consequently, e(P, L) ≤ 15, a contradiction. Hence x2 a4 6∈ E. Thus e(x2 x5 , L) ≤ 5 and so e(x3 x4 , L) ≥ 8. If e(x2 , L) > 0, then [x1 , x2 , P ′ ] ⊇ C5 where P ′ = L − {ai , ai+1 } for some {ai , ai+1 } ⊆ V (L). As H 6⊇ F ⊎ C5 , e(x3 x4 , ai ai+1 ) ≤ 2. Consequently, e(x3 x4 , P ′ ) = 6, e(x3 x4 , ai ai+1 ) = 2 and e(x2 x5 , L) = 5. Hence e(at , x2 x5 ) = 1 for all at ∈ V (L). Thus [P − x1 + aj ] ⊇ F and x1 → (L, aj ) where aj ∈ V (P ′ ) ∩ {a3 , a5 }, a contradiction. Therefore e(x2 , L) = 0 and so e(x3 x4 , L) = 10 and e(x5 , L) = 3. Consequently, H ⊇ 2C5 or H ⊇ F ⊎ C5 , a contradiction. Therefore the three vertices in N (x1 , L) are consecutive on L. Say N (x1 , L) = {a1 , a2 , a3 }. Then I(x2 x5 , L) ⊆ {a1 , a3 } since H 6⊇ 2C5 and H 6⊇ F ⊎ C5 . Thus e(x2 x5 , L) ≤ 7 and so e(x3 x4 , L) ≥ 6. Assume e(x2 , a4 a5 ) ≥ 1. Say w.l.o.g. x2 a4 ∈ E. Then [x1 , x2 , a2 , a3 , a4 ] ⊇ C5 and [x1 , x2 , a1 , a5 , a4 ] ⊇ C5 . As H 6⊇ F ⊎ C5 and by Lemma 2.1(a), e(x3 x4 , a1 a5 ) ≤ 2 and e(x3 x4 , a2 a3 ) ≤ 2. It follows that e(x2 x5 , L) = 7, e(x3 x4 , L) = 6, e(a4 , x3 x4 ) = 2, and e(x2 x5 , a1 a3 ) = 4. Then [x1 , x5 , a1 , a2 , a3 ] ⊇ C5 and [a5 , a4 , x2 , x3 , x4 ] ⊇ F , a contradiction. Hence e(x2 , a4 a5 ) = 0 and so e(x2 , L) ≤ 3. Thus e(x3 x4 , L) ≥ 7. Assume e(x2 , a1 a3 ) ≥ 1. Then [x1 , x2 , a1 , a2 , a3 ] ⊇ C5 . Then e(x3 x4 , a4 a5 ) ≤ 2 as H 6⊇ F ⊎ C5 . Thus e(x3 x4 , a1 a2 a3 ) ≥ 5. As H 6⊇ F ⊎ C5 and x1 → (L, a2 ), we have e(a2 , x2 x4 ) ≤ 1. As e(P, L) ≥ 16, it follows that e(a2 , x2 x4 ) = 1, e(x3 , a1 a2 a3 ) = 3, e(x3 x4 , a4 a5 ) = 2 and e(x5 , L) = 3. As H 6⊇ F ⊎ C5 and x1 → (L, a2 ), we see that x5 a2 6∈ E. Then e(x5 , a4 a5 ) ≥ 1 and so [x3 , x4 , x5 , a4 , a5 ] ⊇ F , a contradiction. Hence e(x2 , a1 a3 ) = 0 and so e(x2 , L) ≤ 1. If e(x5 , L) = 3 then we also have e(x4 , L) ≤ 1 by the symmetry and so e(P, L) ≤ 13, a contradiction. Hence e(x5 , L) ≤ 2. It follows that so e(x3 x4 , L) = 10, e(x2 , L) = 1 and e(x5 , L) = 2. Thus e(a2 , x2 x4 ) = 2 and so H ⊇ F ⊎ C5 , a contradiction. Case 3. e(x1 , L) = 2. Then e(x5 , L) ≤ 2 and e(x3 x4 , L) ≥ 7. First, suppose that the two vertices in N (x1 , L) are not consecutive on L. Say N (x1 , L) = {a1 , a3 }. Assume e(x2 , a1 a3 ) ≥ 1. Then [x1 , x2 , a1 , a2 , a3 ] ⊇ C5 . As H 6⊇ F ⊎ C5 and by Lemma 2.1(a), e(x3 x4 , a4 a5 ) ≤ 2. Hence e(x3 x4 , a1 a2 a3 ) ≥ 5. As x1 → (L, a2 ) and H 6⊇ F ⊎ C5 , e(a2 , x2 x4 ) ≤ 1. As e(P, L) ≥ 16, it follows that e(a2 , x2 x4 ) = 1, e(x5 , L) = 2, e(x2 , L − a2 ) = 4, e(x3 , a1 a2 a3 ) = 3 and

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e(x3 x4 , a4 a5 ) = 2. As [x3 , x4 , x5 , a4 , a5 ] 6⊇ F , e(x5 , a4 a5 ) = 0 by Lemma 2.1(a). As x1 → (L, a2 ) and H 6⊇ F ⊎ C5 , a2 x5 6∈ E. Thus e(x5 , a1 a3 ) = 2. It follows that [x1 , x2 , a1 , a5 , a4 ] ⊇ C5 and [x3 , x4 , x5 , a3 , a2 ] ⊇ C5 , a contradiction. Hence e(x2 , a1 a3 ) = 0. Thus e(x3 x4 , L) ≥ 9. As e(x3 x4 , L) ≤ 10, e(x2 , L) ≥ 2 and so e(x2 , a4 a5 ) ≥ 1. Say w.l.o.g. x2 a4 ∈ E. Then [x1 , x2 , a4 , a5 , a1 ] ⊇ C5 . As H 6⊇ F ⊎ C5 and by Lemma 2.1(a), e(x3 x4 , a2 a3 ) ≤ 2 and so e(x3 x4 , L) ≤ 8, a contradiction. Therefore the two vertices in N (x1 , L) are consecutive on L. Say N (x1 , L) = {a1 , a2 }. Assume x2 a4 ∈ E. Then [x1 , x2 , a4 , a5 , a1 ] ⊇ C5 and [x1 , x2 , a4 , a3 , a2 ] ⊇ C5 . Thus e(x3 x4 , a2 a3 ) ≤ 2 and e(x3 x4 , a1 a5 ) ≤ 2 since H 6⊇ F ⊎ C5 . Hence e(x3 x4 , L) ≤ 6, a contradiction. Hence x2 a4 6∈ E. Thus e(x3 x4 , L) ≥ 8. Assume e(x2 , a3 a5 ) ≥ 1. Say x2 a3 ∈ E. Then [x1 , x2 , a3 , a2 , a1 ] ⊇ C5 and so e(x3 x4 , a4 a5 ) ≤ 2. It follows that e(x3 x4 , a1 a2 a3 ) = 6, e(x3 x4 , a4 a5 ) = 2, e(x2 , L − a4 ) = 4 and e(x5 , L) = 2. As x2 a5 ∈ E and by the symmetry, we also have e(x3 x4 , a5 a1 a2 ) = 6. Then H ⊇ F ⊎ C5 , a contradiction. Therefore e(x2 , a3 a5 ) = 0. It follows that e(x2 , a1 a2 ) = 2, e(x3 x4 , L) = 10 and e(x5 , L) = 2. Then H ⊇ F ⊎ C5 , a contradiction Lemma 2.4. Let D and L be disjoint subgraphs of G with D ∼ = C5 . = F2 and L ∼ Let R be the set of the three vertices of D with degree 2 in D. If e(R, L) ≥ 10, then [D, L] ⊇ F1 ⊎ C5 . Proof. As e(R, L) ≥ 10, e(u, L) ≥ 4 for some u ∈ R. Thus u → (L, v) for some v ∈ V (L) with e(v, R − {u}) ≥ 1. Clearly, [D − u + v] ⊇ F1 . ∼ F and L ∼ Lemma 2.5. Let D and L be disjoint subgraphs of G with D = = C5 . Suppose that {D, L} is optimal and e(D, L) ≥ 16. Then [D, L] contains one of F1 ⊎ C5 , F2 ⊎ C5 , B ⊎ C5 and 2C5 , or there exist two labellings D = x0 x1 x2 x3 x4 x1 and L = a1 a2 a3 a4 a5 a1 such that e(x0 , L) = 0, e(x1 x3 , L) = 10, N (x2 , L) = N (x4 , L) = {a1 , a2 , a4 }, τ (L) = 4 and a3 a5 6∈ E. Proof. Say H = [D, L]. Suppose that H does not contain any of F1 ⊎ C5 , F2 ⊎ C5 , B ⊎ C5 and 2C5 . We shall prove that there exist two labellings of D and L satisfying the property in the lemma. Say D = x0 x1 x2 x3 x4 x1 and L = a1 a2 a3 a4 a5 a1 . Then x2 x4 6∈ E. Let Q = x1 x2 x3 x4 x1 . If e(x0 , L) ≥ 4, then for each ai ∈ V (L), [L−ai +x0 ] ⊇ C5 or [L−ai +x0 ] ⊇ F1 . Thus [Q+ai ] 6⊇ C5 and so e(ai , Q) ≤ 2 for each ai ∈ V (L). Consequently, e(D, L) ≤ 15, a contradiction. Therefore e(x0 , L) ≤ 3. We divide the proof into the following cases. Case 1. e(x0 , L) = 0. First, suppose that e(x2 , L) ≥ 4 or e(x4 , L) ≥ 4. Say, {a1 , a2 , a3 , a4 } ⊆ N (x2 , L). Assume e(x1 , a2 a3 ) ≥ 1. Say w.l.o.g. x1 a2 ∈ E. Then [x0 , x1 , x2 , a2 , a1 ] ⊇ F1 and [x0 , x1 , x2 , a2 , a3 ] ⊇ F1 . As H 6⊇ F1 ⊎ C5 , we see that e(x3 x4 , a3 a5 ) ≤ 2 and e(x3 x4 , a1 a4 ) ≤ 2. As e(Q, L) ≥ 16, it follows that e(x1 x2 , L) = 10 and e(a2 , x3 x4 ) = 2. Thus [x0 , x1 , a2 , x3 , x4 ] ⊇ F1 and x2 →

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(L, a2 ), a contradiction. Hence e(x1 , a2 a3 ) = 0. As e(x1 , L) ≥ 1, this argument implies that e(x2 , L) 6= 5. Similarly, e(x4 , L) 6= 5. As e(Q, L) ≥ 16, it follows that e(x1 , a1 a5 a4 ) = 3, e(x3 , L) = 5 and e(x4 , L) = 4. Then [x0 , x1 , x2 , a1 , a2 ] ⊇ F1 and [x3 , x4 , a3 , a4 , a5 ] ⊇ C5 , a contradiction. Hence e(x2 , L) ≤ 3 and e(x4 , L) ≤ 3. Consequently, e(x1 x3 , L) = 10, e(x2 , L) = e(x4 , L) = 3. Then x2 is adjacent two consecutive vertices of L. Say w.l.o.g. e(x2 , a1 a2 ) = 2. Then [x0 , x1 , x2 , a1 , a2 ] ⊇ F1 . Thus e(x4 , a3 a5 ) = 0 as H 6⊇ F1 ⊎ C5 . Hence e(x4 , a1 a2 a4 ) = 3. Similarly, e(x2 , a1 a2 a4 ) = 3. Clearly, [D − x3 + ai ] ⊇ F for i ∈ {1, 2}. As {D, L} is na optimal, x3 → (L, ai ) for i ∈ {1, 2}. This implies that τ (a1 , L) = τ (a2 , L) = 2. As [x0 , x1 , x2 , a1 , a2 ] ⊇ F1 , [x3 , x4 , a3 , a4 , a5 ] 6⊇ C5 . This implies that a3 a5 6∈ E. Therefore these two labellings satisfy the property described in the lemma. Case 2. e(x0 , L) = 1. Then e(Q, L) ≥ 15. Say x0 a1 ∈ E. First, suppose e(x1 , a3 a4 ) ≥ 1. Say w.l.o.g. x1 a3 ∈ E. Then [x1 , x0 , a1 , a2 , a3 ] ⊇ C5 . By Lemma 2.1(c), we have e(a4 a5 , x2 x3 x4 ) ≤ 3 since H 6⊇ 2C5 , H 6⊇ F1 ⊎ C5 and H 6⊇ F2 ⊎ C5 . Thus e(a4 a5 , Q) ≤ 5. Similarly, if x1 a4 ∈ E then e(a2 a3 , Q) ≤ 5 and so e(Q, L) ≤ 14, a contradiction. Hence x1 a4 6∈ E. Thus e(a4 a5 , Q) ≤ 4 and so e(a1 a2 a3 , Q) ≥ 11. This implies that if e(a2 , x1 x3 ) = 2 then there is a choice {i, j} = {2, 4} such that e(xi , a1 a3 ) = 2 and e(a2 , x1 xj x3 ) = 3. Thus [x0 , x1 , xj , x3 , a2 ] ⊇ F1 and xi → (L, a2 ), a contradiction. Hence e(a2 , x1 x3 ) = 1, e(a1 a3 , Q) = 8, e(a2 , x2 x4 ) = 2 and e(a4 a5 , Q) = 4 with a5 x1 ∈ E. Consequently, [a4 , a5 , a1 , x0 , x1 ] ⊇ F1 and [a2 , a3 , x2 , x3 , x4 ] ⊇ C5 , a contradiction. Therefore e(x1 , a3 a4 ) = 0. Next, suppose e(x1 , a1 a5 ) = 2 or e(x1 , a1 a2 ) = 2. Say w.l.o.g. e(x1 , a1 a5 ) = 2. Then [a4 , a5 , a1 , x0 , x1 ] ⊇ F1 . Thus e(a2 a3 , x2 x4 ) ≤ 2. Hence e(a2 a3 , Q) ≤ 5 and so e(a1 a5 a4 , x2 x3 x4 ) ≥ 8. This implies that if x3 a5 ∈ E then there is a choice {i, j} = {2, 4} such that e(a5 , x1 xi x3 ) = 3, e(xj , a1 a4 ) = 2 and consequently, H ⊇ F1 ⊎C5 , a contradiction. Hence a5 x3 6∈ E and it follows that e(a1 , x2 x3 x4 ) = 3, e(a5 , x2 x4 ) = 2, e(a4 , x2 x3 x4 ) = 3, e(a2 a3 , Q) = 5 with a2 x1 ∈ E. Then [a3 , a2 , a1 , x0 , x1 ] ⊇ F1 and [a4 , a5 , x2 , x3 , x4 ] ⊇ C5 , a contradiction. Therefore e(x1 , a1 a5 ) ≤ 1 and e(x1 , a1 a2 ) ≤ 1. Thus e(x1 , L) ≤ 2. Assume that a1 x3 ∈ E. Then x2 6→ (L, a1 ) as H 6⊇ 2C5 . Hence e(x2 , a2 a5 ) ≤ 1, and similarly, e(x4 , a2 a5 ) ≤ 1. As e(Q, L) ≥ 15, it follows that e(x1 , a2 a5 ) = 2, e(x3 , L) = 5, e(x2 x4 , a1 a3 a4 ) = 6 and e(x2 , a2 a5 ) = e(x4 , a2 a5 ) = 1. Say w.l.o.g. a5 x4 ∈ E. Then [D−x2 +a5 ] ⊇ F1 and x2 → (L, a5 ), a contradiction. Therefore a1 x3 6∈ E. If x1 a1 ∈ E then e(x1 , a2 a5 ) = 0 and so e(a1 , Q − x3 ) + e(L − a1 , Q − x1 ) ≥ 15. Then [D − x2 + a1 ] ⊇ F1 and x2 → (L, a1 ), a contradiction. Hence N (x1 , L) ⊆ {a2 , a5 }. As e(Q, L) ≥ 15, e(a2 a5 , x2 x4 ) ≥ 3 and e(a2 a4 , x3 xi ) ≥ 3 for i ∈ {2, 4}. Say w.l.o.g. x2 a5 ∈ E. Then [x0 , x1 , x2 , a5 , a1 ] ⊇ C5 and [x3 , x4 , a2 , a3 , a4 ] ⊇ C5 , a contradiction. Case 3. N (x0 , L) = {ai , ai+2 } for some i ∈ {1, 2, 3, 4, 5}. Say N (x0 , L) = {a1 , a3 }. Then e(Q, L) ≥ 14. As H 6⊇ 2C5 , e(a2 , Q) ≤ 2. We claim that

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e(x1 , a1 a3 ) = 0. On the contrary, say e(x1 , a1 a3 ) ≥ 1. Then [x0 , x1 , a1 , a2 , a3 ] ⊇ C5 . Since H 6⊇ 2C5 , H 6⊇ F1 ⊎C5 and H 6⊇ F2 ⊎C5 , we see that e(a4 a5 , x2 x3 x4 ) ≤ 3 by Lemma 2.1(c). Thus e(a4 a5 , Q) ≤ 5 and e(a1 a3 , Q) ≥ 14 − e(a2 , Q) − e(a4 a5 , Q) ≥ 7. As e(a1 a3 , Q) ≤ 8, it follows that either e(a1 , Q) = 4 and x1 a5 ∈ E or e(a3 , Q) = 4 and x1 a4 ∈ E. Say w.l.o.g. the former holds. Then [D−x3 +a1 ] ⊇ F2 , [x0 , x1 , a1 , a5 , a4 ] ⊇ F1 and [x0 , x1 , a1 , a5 , xi ] ⊇ F2 for i ∈ {2, 4}. Furthermore, if x1 a2 ∈ E then [x0 , x1 , a1 , a5 , a2 ] ⊇ F2 and [x0 , x1 , a1 , a2 , xi ] ⊇ F2 for i ∈ {2, 4}. Assume for the moment that e(a3 , x2 x4 ) = 2. Then we see that e(a2 , x2 x4 ) = 0 as H 6⊇ F1 ⊎ C5 . If x1 a2 ∈ E, then e(a4 , x2 x4 ) = 0 as H 6⊇ F2 ⊎ C5 and for the same reason, [a3 , a4 , a5 , x3 , xi ] 6⊇ C5 for i ∈ {2, 4}. This implies that x3 a5 6∈ E and so e(a5 , x2 x4 ) ≥ 1 since 8 ≥ e(a1 a3 , Q) ≥ 14 − e(a2 , Q) − e(a4 a5 , Q) ≥ 7. Thus x3 a3 6∈ E since [a3 , a4 , a5 , x3 , xi ] 6⊇ C5 for i ∈ {2, 4}. It follows that {a3 x1 , x3 a4 } ⊆ E. Consequently, [a1 , a5 , a4 , x2 , x3 ] ⊇ C5 and [x0 , x1 , x4 , a2 , a3 ] ⊇ F2 , a contradiction. Hence x1 a2 6∈ E. As e(Q, L) ≥ 14, it follows that a2 x3 ∈ E, e(a1 a3 , Q) = 8, e(x1 , a4 a5 ) = 2 and e(a4 a5 , x2 x3 x4 ) = 3. Say w.l.o.g. a4 x2 ∈ E. Then [a2 , a3 , a4 , x2 , x3 ] ⊇ C5 and so H ⊇ F2 ⊎ C5 , a contradiction. Hence e(a3 , x2 x4 ) ≤ 1. It follows that e(a3 , x2 x4 ) = 1, e(a3 , x1 x3 ) = 2, e(a2 , Q) = 2 and e(a4 a5 , Q) = 5 with e(x1 , a4 a5 ) = 2. Thus [x0 , x1 , a5 , a4 , a3 ] ⊇ C5 and so e(a2 , x1 x3 ) = 2 as H 6⊇ 2C5 . Say w.l.o.g. a3 x2 ∈ E. As H 6⊇ F2 ⊎ C5 , we see that [x2 , x3 , a5 , a4 , a3 ] 6⊇ C5 and [a3 , a4 , x2 , x3 , x4 ] 6⊇ C5 . This implies that e(a5 , x2 x3 ) = 0 and a4 x4 6∈ E. As e(a4 a5 , x2 x3 x4 ) = 3, it follows that [a4 , a5 , x2 , x3 , x4 ] ⊇ C5 and so H ⊇ 2C5 , a contradiction. Therefore e(x1 , a1 a3 ) = 0. Assume e(x1 , a4 a5 ) = 0. As e(Q, L) ≥ 14, it follows that e(x2 x3 x4 , L−a2 ) = 12 and e(a2 , Q) = 2. Thus [x2 , x3 , x4 , a4 , a5 ] ⊇ K5− . As [x1 , x0 , a1 , a2 , a3 ] ⊇ F , we have τ (L) ≥ 4 by the optimality of {D, L}. Consequently, x0 → (L, ar ) for some r ∈ {4, 5} and so H ⊇ 2C5 as [Q + ar ] ⊇ C5 , a contradiction. Hence e(x1 , a4 a5 ) ≥ 1. Say w.l.o.g. x1 a5 ∈ E. Then [x0 , x1 , a5 , a4 , a3 ] ⊇ C5 . Since H 6⊇ 2C5 , H 6⊇ F1 ⊎ C5 and H 6⊇ F2 ⊎ C5 , we see that e(a1 a2 , x2 x3 x4 ) ≤ 3 by Lemma 2.1(c). Thus e(a1 a2 , Q) ≤ 4 and so e(a3 a4 a5 , Q) ≥ 10. Hence e(a4 a5 , Q) ≥ 7. As above, we shall have that [x2 , x3 , x4 , a4 , a5 ] 6⊇ K5− . This implies that e(a4 a5 , x2 x3 x4 ) 6= 6. Thus e(a4 a5 , x2 x3 x4 ) = 5, e(x1 , a4 a5 ) = 2, e(a3 , x2 x3 x4 ) = 3 and e(a1 a2 , Q) = 4. Similarly, we shall have e(a1 , x2 x3 x4 ) = 3 as [x0 , x1 , a4 , a5 , a1 ] ⊇ C5 . As e(a4 a5 , x2 x3 x4 ) = 5, we may assume w.l.o.g. that e(a4 , x2 x3 x4 ) = 3. Thus [a3 , a4 , x2 , x3 , x4 ] ⊇ K5− and [a2 , a1 , a5 , x1 , x0 ] ⊇ F . By the optimality of {D, L}, we shall have τ (L) ≥ 4. Thus x0 → (L, ar ) for some r ∈ {4, 5} and so H ⊇ 2C5 , a contradiction. Case 4. N (x0 , L) = {ai , ai+1 } for some i ∈ {1, 2, 3, 4, 5}. Say, N (x0 , L) = {a1 , a2 }. First, suppose that x1 a4 ∈ E. Then [x0 , x1 , a4 , a5 , a1 ] ⊇ C5 and [x0 , x1 , a4 , a3 , a2 ] ⊇ C5 . Since H 6⊇ 2C5 , H 6⊇ F1 ⊎ C5 and H 6⊇ F2 ⊎ C5 , we see that e(a2 a3 , Q − x1 ) ≤ 3 and e(a1 a5 , Q − x1 ) ≤ 3 by Lemma 2.1(c). As e(Q, L) ≥ 14, it follows that e(x1 , L) = 5, e(a4 , Q) = 4, e(a2 a3 , Q − x1 ) = 3 and

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e(a1 a5 , Q − x1 ) = 3. Then [x0 , x1 , a5 , a1 , a2 ] ⊇ C5 and so e(a3 a4 , Q − x1 ) ≤ 3. Thus e(a3 , Q − x1 ) = 0 as e(a4 , Q − x1 ) = 3. Similarly, e(a5 , Q − x1 ) = 0. Thus e(a1 a2 , Q − x1 ) = 6. Then [a1 , x2 , x3 , a4 , a5 ] ⊇ C5 and [a3 , a2 , x0 , x1 , x4 ] ⊇ F2 , a contradiction. Hence x1 a4 6∈ E. Next, suppose e(x3 , a1 a2 ) = 2. Then e(xi , a1 a3 ) ≤ 1 and e(xi , a2 a5 ) ≤ 1 for each i ∈ {2, 4} as H 6⊇ 2C5 . Thus e(x2 x4 , L−a4 ) ≤ 4 and so e(x1 , L−a4 )+e(x3 , L)+ e(a4 , x2 x4 ) ≥ 10. Then e(x1 , a1 a2 ) ≥ 1. Thus [xi , x1 , x0 , a1 , a2 ] ⊇ F1 for i ∈ {2, 4}. Clearly, e(x3 , a3 a5 ) ≥ 1. Assume e(x3 , a3 a5 ) = 2. Then e(x2 x4 , a3 a5 ) = 0 as H 6⊇ F1 ⊎ C5 . If e(a4 , x2 x4 ) = 1, then e(x1 , L − a4 ) = 4 , e(x3 , L) = 5 and e(x2 x4 , a1 a2 ) = 4. Thus [x0 , x1 , x4 , a2 , a3 ] ⊇ F2 and [x3 , a4 , a5 , a1 , x2 ] ⊇ C5 , a contradiction. Hence e(a4 , x2 x4 ) = 2. If x3 a4 ∈ E then [x2 , x3 , x4 , a4 , ai ] ⊇ F2 for i ∈ {3, 5}. As e(x1 , a3 a5 ) ≥ 1, we see that H ⊇ F2 ⊎ C5 , a contradiction. Thus x3 a4 6∈ E, e(x1 , L−a4 ) = 4, e(x3 , L−a4 ) = 4, e(a4 , x2 x4 ) = 2 and e(x2 x4 , a1 a2 ) = 4. Thus [x0 , x1 , x4 , a2 , a3 ] ⊇ F2 and [x3 , a1 , a5 , a4 , x2 ] ⊇ C5 , a contradiction. We conclude that e(x3 , a3 a5 ) = 1. Thus e(x1 , L − a4 ) = 4, e(x3 , L) = 4 and e(a4 , x2 x4 ) = 2. Say w.l.o.g. x3 a5 ∈ E. Then [x2 , x4 , a5 , a4 , x3 ] ⊇ F2 and [x0 , x1 , a1 , a2 , a3 ] ⊇ C5 , a contradiction. Therefore e(x3 , a1 a2 ) ≤ 1. Next, suppose that e(x2 , a1 a2 ) ≥ 1 and e(x4 , a1 a2 ) ≥ 1. Then [xi , x1 , x0 , a1 , a2 ] ⊇ C5 for i ∈ {2, 4}. Since H 6⊇ 2C5 , H 6⊇ F1 ⊎ C5 and H 6⊇ F2 ⊎ C5 , we see that e(x3 xi , a3 a4 a5 ) ≤ 3 for i ∈ {2, 4} by Lemma 2.1(c). Furthermore, if for some i ∈ {2, 4}, say i = 2, we have e(x2 , a3 a4 a5 ) = 3, then [x2 , a3 , a4 , a5 , aj ] ⊇ F1 for j ∈ {1, 2} and so e(x3 , a1 a2 ) = 0 since H 6⊇ C5 ⊎ F1 . Consequently, e(x1 , L − a4 ) = 4, e(x2 x4 , L) = 10 and so H ⊇ 2C5 , a contradiction. Therefore if e(x3 , a3 a4 a5 ) = 0 then e(xi , a3 a4 a5 ) ≤ 2 for i ∈ {2, 4}. Together with x1 a4 6∈ E and e(x3 , a1 a2 ) ≤ 1, we see that if e(x3 , a3 a4 a5 ) = 0 or e(x3 , a3 a4 a5 ) > 1 then e(Q, L) ≤ 13, a contradiction. Hence e(x3 , a3 a4 a5 ) = 1. It follows that e(x1 , L − a4 ) = 4, e(x3 , a1 a2 ) = 1, e(x2 x4 , a1 a2 ) = 4, e(x2 , a3 a4 a5 ) = 2 and e(x4 , a3 a4 a5 ) = 2. If e(x3 , a3 a5 ) = 1, then either [x2 , x3 , a3 , a4 , a5 ] ⊇ C5 or [x2 , x3 , a3 , a4 , a5 ] ⊇ F1 , and consequently, H ⊇ C5 ⊎ F1 , a contradiction. Hence x3 a4 ∈ E. Then we see that [x2 , x3 , a4 , a5 , a1 ] ⊇ C5 and [x0 , x1 , x4 , a2 , a3 ] ⊇ F2 , a contradiction. Therefore either e(x2 , a1 a2 ) = 0 or e(x4 , a1 a2 ) = 0. Say w.l.o.g. e(x4 , a1 a2 ) = 0. Finally, if e(x2 , a1 a2 ) ≥ 1 then, as above, we would have e(x3 x4 , a3 a4 a5 ) ≤ 3 and so e(Q, L) ≤ 13, a contradiction. Hence e(x2 , a1 a2 ) = 0. As e(Q, L) ≥ 14, it follows that e(x1 , L − a4 ) = 4, e(x3 , L − ai ) = 4 for some i ∈ {1, 2} and e(x2 x4 , a3 a4 a5 ) = 6. As [x2 , x3 , x4 , a4 , a5 ] ⊇ C5 , we see H ⊇ 2C5 , a contradiction. Case 5. N (x0 , L) = {ai , ai+1 , ai+2 } for some i ∈ {1, 2, 3, 4, 5}. Say N (x0 , L) = {a1 , a2 , a3 }. Then for each i ∈ {2, 4, 5}, [L − ai + x0 ] ⊇ C5 or [L − ai + x0 ] ⊇ F1 and so e(ai , Q) ≤ 2. Thus e(a1 a3 , Q) ≥ 7. Hence [Q + ai ] ⊇ C5 for each i ∈ {1, 3}. Therefore [L − ai + x0 ] 6⊇ C5 and [L − ai + x0 ] 6⊇ B for each i ∈ {1, 3}. This implies that τ (L) ≤ 1. As e(a1 a3 , Q) ≤ 8, e(a4 a5 , Q) ≥ 3. Say

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w.l.o.g. e(a5 , Q) = 2. As [Q + a5 ] 6⊇ C5 , N (a5 , Q) = {x2 , x4 } or N (a5 , Q) = {x1 , x3 }. First, assume N (a5 , Q) = {x2 , x4 }. Then [a4 , a5 , x2 , x3 , x4 ] ⊇ F . As e(a1 a3 , Q) ≥ 7, e(x1 , a1 a3 ) ≥ 1 and so [x0 , x1 , a1 , a2 , a3 ] ⊇ C ′ ∼ = C5 with τ (C ′ ) ≥ 2, contradicting the optimality of {D, L}. Hence N (a5 , Q) = {x1 , x3 }. Then [a4 , a5 , x1 , xi , x3 ] ⊇ F for each i ∈ {2, 4}. By the optimality of {D, L} and Lemma 2.1(b), we get e(xi , a1 a3 ) ≤ 1 for each i ∈ {2, 4} and so e(a1 a3 , Q) ≤ 6, a contradiction. Case 6. N (x0 , L) = {ai , ai+1 , ai+3 } for some i ∈ {1, 2, 3, 4, 5}. Say N (x0 , L) = {a1 , a2 , a4 }. Clearly, x0 → (L, a3 ) and x0 → (L, a5 ). Thus e(a3 , Q) ≤ 2 and e(a5 , Q) ≤ 2 for otherwise H ⊇ 2C5 . As H 6⊇ 2C5 , we see that x0 6→ L and so a3 a5 6∈ E. As e(Q, L) ≥ 13, e(a3 a5 , Q) ≥ 1. Say w.l.o.g. e(a5 , Q) ≥ 1. Then [Q + a5 ] ⊇ F . By the optimality of {D, L}, τ (L) ≥ τ (x0 a1 a2 a3 a4 x0 ). This implies that a2 a5 ∈ E. Similarly, if e(a3 , Q) ≥ 1 then a1 a3 ∈ E. Assume a1 a3 6∈ E. Then e(a3 , Q) = 0 and so e(a1 a2 a4 , Q) ≥ 11. Then e(ar , Q) = 4 for some r ∈ {1, 2} and [L − ar + x0 ] ⊇ F . As τ (ar x1 x2 x3 x4 ar ) ≥ 3, it follows that τ (L) = 3 and so {a1 a4 , a2 a4 } ⊆ E. Thus [L − a1 + x0 ] ⊇ F2 and [Q + a1 ] ⊇ C5 , a contradiction. Therefore a1 a3 ∈ E. Thus [L − a4 + x0 ] ⊇ F2 . Hence [Q + a4 ] 6⊇ C5 and so e(a4 , Q) ≤ 2. Consequently, e(a1 a2 , Q) ≥ 7 and so [Q + ai ] ⊇ C5 for each i ∈ {1, 2}. Hence a1 a4 6∈ E and a2 a4 6∈ E for otherwise H ⊇ F2 ⊎ C5 . Hence τ (L) = 2. By the optimality of {D, L}, [Q + ai ] 6⊇ C with C∼ = C5 and τ (C) ≥ 3 for each i ∈ {1, 2}. This implies that e(ai , Q) ≤ 3 for each i ∈ {1, 2} and therefore e(a1 a2 , Q) ≤ 6, a contradiction. Lemma 2.6. Let D, L1 and L2 be disjoint subgraphs of G with D ∼ = F and L1 ∼ = L2 ∼ = C5 . Suppose that L1 = a1 a2 a3 a4 a5 a1 , V (D) = {x0 , x1 , x2 , x3 , x4 } and E(D) = {x0 x1 , x1 x2 , x2 x3 , x3 x4 , x4 x1 } such that e(x0 , L1 ) = 0, and e(x1 x3 , L1 ) = 10, N (x2 , L1 ) = N (x4 , L1 ) = {a1 , a2 , a4 }, τ (L1 ) = 4 and a3 a5 6∈ E. Suppose that e(x0 x2 a3 a5 , L2 ) ≥ 13. Then [D, L1 , L2 ] contains either of F1 ⊎ 2C5 or 3C5 . Proof. For the proof, we may assume that none of x0 x3 , x1 x3 and x2 x4 is an edge as they will not be used in the proof. Set G1 = [D, L1 ], G2 = [G1 , L2 ] and R = {x0 , x2 , a3 , a5 }. It is easy to see that for any permutation f of {x2 , a3 , a5 }, we can extend f to be an automorphism of G1 such that every vertex of G1 − {x2 , a3 , a5 } is fixed under f . Therefore x2 , a3 and a5 are in the symmetric position in the following argument. On the contrary, suppose that G2 6⊇ F1 ⊎ 2C5 and G2 6⊇ 3C5 . It is easy to check that if u → (L2 ; R − {u}) for some u ∈ R then G2 ⊇ F1 ⊎ 2C5 or G2 ⊇ 3C5 . Therefore u 6→ (L2 ; R − {u}) for each u ∈ R. By Lemma 2.1(d), there exist two labellings R = {y1 , y2 , y3 , y4 } and L2 = b1 b2 b3 b4 b5 b1 such that e(y1 y2 , b1 b2 b3 b4 ) = 8, e(y3 , b1 b5 b4 ) = 3 and e(y4 , b1 b4 ) = 2. If x0 ∈ {y1 , y2 }, we may assume that {y1 , y2 } = {x0 , x2 }. Then [x0 , x1 , x2 , b2 , b3 ] ⊇ C5 , [a3 , a5 , b1 , b5 , b4 ] ⊇ C5 and [x3 , x4 , a1 , a2 , a4 ] ⊇ C5 , a contradiction. Hence x0 6∈

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{y1 , y2 }. Say w.l.o.g. that {y1 , y2 } = {a3 , a5 }. Thus [a3 , a4 , a5 , b2 , b3 ] ⊇ C5 , [x0 , x2 , b1 , b5 , b4 ] ⊇ C5 and [x1 , x4 , x3 , a1 , a2 ] ⊇ C5 , a contradiction. Lemma 2.7. Let D and L be disjoint subgraphs of G with D ∼ = K4+ and L ∼ = B. Let R be the set of the four vertices of L with degree 2 in L. Suppose that e(D, R) ≥ 13. Then either [D, L] ⊇ K4+ ⊎ C5 or [D, L] ⊇ 2C5 or [D, L] ⊇ B ⊎ C5 . Proof. Say H = [D, L]. On the contrary, suppose that H contains none of K4+ ⊎ C5 , 2C5 and B ⊎ C5 . Say V (D) = {x0 , x1 , x2 , x3 , x4 } with e(x0 , D) = 1 and x0 x1 ∈ E. Let Q = [x1 , x2 , x3 , x4 ]. Say L = a0 a1 a2 a0 a3 a4 a0 . Then Q ∼ = K4 and R = {a1 , a2 , a3 , a4 }. If e(x0 , R) ≥ 3, say w.l.o.g. e(x0 , a1 a2 a3 ) = 3, then [L − ai + x0 ] ⊇ C5 and so Q + ai 6⊇ C5 for each i ∈ {1, 2, 4}. Consequently, e(ai , Q) ≤ 1 for all i ∈ {1, 2, 4} and so e(D, R) ≤ 11, a contradiction. Hence e(x0 , R) ≤ 2. Suppose that e(x0 , R) = 2. Then e(R, Q) ≥ 11. First, assume e(x0 , a1 a2 ) = 1 and e(x0 , a3 a4 ) = 1. Say w.l.o.g. e(x0 , a1 a3 ) = 2. Then e(a2 , Q) ≤ 1 and e(a4 , Q) ≤ 1 as H 6⊇ 2C5 . Consequently, e(R, Q) ≤ 10, a contradiction. Therefore we may assume w.l.o.g. that e(x0 , a1 a2 ) = 2. We claim e(x1 , a1 a2 ) = 0. To see this, suppose e(x1 , a1 a2 ) ≥ 1. Then [x0 , x1 , a1 , a2 , a0 ] ⊇ C5 . Thus e(a3 a4 , x2 x3 x4 ) ≤ 2 for otherwise [a3 , a4 , x2 , x3 , x4 ] ⊇ C5 or [a3 , a4 , x2 , x3 , x4 ] ⊇ K4+ . Thus e(a3 a4 , Q) ≤ 4 and so e(a1 a2 , Q) ≥ 7. Say w.l.o.g. e(a1 , Q) = 4. Then [D − xi + a1 ] ⊇ K4+ for each i ∈ {2, 3, 4} and so [L − a1 + xi ] 6⊇ C5 for each i ∈ {2, 3, 4}. Thus I(a2 a3 , Q − x1 ) = ∅ and so e(a2 a3 , Q) ≤ 5. Hence e(a4 , Q) ≥ 2. Similarly, e(a3 , Q) ≥ 2. It follows that [a3 , a4 , x2 , x3 , x4 ] ⊇ C5 or [a3 , a4 , x2 , x3 , x4 ] ⊇ B, a contradiction. This shows that e(x1 , a1 a2 ) = 0. Suppose e(a1 , Q − x1 ) = 3 or e(a2 , Q − x1 ) = 3. Then [x0 , x1 , xi , a1 , a2 ] ⊇ C5 for each i ∈ {2, 3, 4}. Thus [xi , xj , a0 , a3 , a4 ] 6⊇ C5 and [xi , xj , a0 , a3 , a4 ] 6⊇ B for each 2 ≤ i < j ≤ 4. This implies that e(a3 a4 , Q − x1 ) ≤ 2. Hence e(a1 a2 , Q) ≥ 7 and so e(x1 , a1 a2 ) ≥ 1, a contradiction. Hence e(ai , Q − x1 ) ≤ 2 for each i ∈ {1, 2} and so e(a3 a4 , Q) ≥ 7. Say w.l.o.g. e(a4 , Q) = 4. Then [D − xi + a4 ] ⊇ K4+ for each i ∈ {2, 3, 4} and therefore I(a1 a3 , Q − x1 ) = ∅ as H 6⊇ K4+ ⊎ C5 . Thus e(a1 a3 , Q) ≤ 4 and so e(a2 , Q) ≥ 3, a contradiction. Next, suppose e(x0 , R) = 1. Then e(Q, R) ≥ 12. Say x0 a1 ∈ E. Suppose e(x1 , a1 a2 ) ≥ 1. Then [x0 , x1 , a1 , a2 , a0 ] ⊇ C5 or [x0 , x1 , a1 , a2 , a0 ] ⊇ B. Thus [x2 , x3 , x4 , a3 , a4 ] 6⊇ C5 . This implies that e(a3 a4 , Q−x1 ) ≤ 3. Thus e(a3 a4 , Q) ≤ 5 and so e(a1 a2 , Q) ≥ 7. Thus [D − xi + a1 ] ⊇ C5 for all i ∈ {2, 3, 4}. As H 6⊇ 2C5 , I(a2 a3 , Q − x1 ) = ∅ and I(a2 a4 , Q − x1 ) = ∅. Hence e(a2 a3 , Q) ≤ 5 and so e(a4 , Q) ≥ 3. Then I(a2 a4 , Q − x1 ) 6= ∅, a contradiction. Hence e(x1 , a1 a2 ) = 0. Thus e(a1 a2 , Q) ≤ 6 and e(a3 a4 , Q) ≥ 6. Then [xi , xj , a3 , a4 , a0 ] ⊇ C5 for some 2 ≤ i < j ≤ 4. Say {i, j, k} = {2, 3, 4}. Then a2 xk 6∈ E as H 6⊇ 2C5 . Therefore e(a1 a2 , Q) ≤ 5 and so e(a3 a4 , Q) ≥ 7. Thus [xr , xt , a3 , a4 , a0 ] ⊇ C5 for all 2 ≤ r < t ≤ 4. Therefore e(a2 , Q − x1 ) = 0 as H 6⊇ 2C5 . Consequently, e(Q, R) ≤ 11, a contradiction. Finally, suppose e(x0 , R) = 0. As e(R, Q) ≥ 13, e(ai , Q) = 4 for some ai ∈ R.

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Say e(a1 , Q) = 4. Then I(a2 a3 , Q − x1 ) = ∅ as H 6⊇ K4+ ⊎ C5 . Thus e(a4 , Q) = 4 as e(R, Q) ≥ 13. Similarly, e(a3 , Q) = 4. Then we readily see that H ⊇ K4+ ⊎ C5 , a contradiction. Lemma 2.8. Let B1 and B2 be disjoint subgraphs of G such that B1 ∼ = B and B2 ∼ = B. Let R be the set of the four vertices of B1 with degree 2 in B1 . Suppose that e(R, B2 ) ≥ 13. Then [B1 , B2 ] ⊇ 2C5 or [B1 , B2 ] ⊇ B ⊎ C5 . Proof. On the contrary, suppose that [B1 , B2 ] 6⊇ 2C5 and [B1 , B2 ] 6⊇ B ⊎ C5 . Say B1 = a0 a1 a2 a0 a3 a4 a0 and B2 = b0 b1 b2 b0 b3 b4 b0 . Then R = {a1 , a2 , a3 , a4 } and e(R, B2 − b0 ) ≥ 9. This implies that e(ai ai+1 , bj bj+1 ) ≥ 3 for some i ∈ {1, 3} and j ∈ {1, 3}. Say w.l.o.g. e(a1 a2 , b1 b2 ) ≥ 3. Then [a1 , a2 , b0 , b1 , b2 ] ⊇ C5 and [b1 , b2 , a0 , a1 , a2 ] ⊇ C5 . Therefore [a0 , a3 , a4 , b3 , b4 ] 6⊇ C5 , [a0 , a3 , a4 , b3 , b4 ] 6⊇ B, [b0 , b3 , b4 , a3 , a4 ] 6⊇ C5 and [b0 , b3 , b4 , a3 , a4 ] 6⊇ B. This implies that e(a3 a4 , b3 b4 ) ≤ 1 and e(b0 , a3 a4 ) ≤ 1. If e(a1 a2 , b3 b4 ) ≥ 3, then we also have that e(a3 a4 , b1 b2 ) ≤ 1 and it follows that e(a1 a2 , B2 ) = 10 and e(a3 a4 , b3 b4 ) = 1 as e(R, B2 ) ≥ 13. Consequently, [B2 − br + a1 ] ⊇ C5 and [B1 − a1 + br ] ⊇ C5 where r ∈ {3, 4} with e(br , a3 a4 ) = 1, a contradiction. Hence e(a1 a2 , b3 b4 ) ≤ 2. Suppose e(a3 a4 , b1 b2 ) ≥ 3. Similarly, we shall have e(a1 a2 , b3 b4 ) ≤ 1, e(b0 , a1 a2 ) ≤ 1 and so e(R, B2 ) ≤ 12, a contradiction. Therefore, e(a3 a4 , b1 b2 ) ≤ 2. Thus e(a3 a4 , B2 ) ≤ 4 and so e(a1 a2 , B2 ) ≥ 9. Consequently, e(a1 a2 , b3 b4 ) ≥ 3, a contradiction. Lemma 2.9. Let D and L be disjoint subgraphs of G with D ∼ = F1 and L ∼ = C5 . Suppose that {D, L} is optimal and e(D, L) ≥ 16. Then [D, L] contains one of K4+ ⊎C5 , K4+ ⊎B, 2C5 and B ⊎C5 , or there exist two labellings L = a1 a2 a3 a4 a5 a1 and V (D) = {x0 , x1 , x2 , x3 , x4 } with E(D) = {x0 x1 , x1 x2 , x2 x3 , x3 x4 , x4 x1 , x2 x4 } such that e(x0 , L) = 0, e(a1 a2 a4 , D − x0 ) = 12, N (a3 , D) = N (a5 , D) = {x2 , x4 }, τ (L) = 4 and a3 a5 6∈ E. Proof. Say H = [D, L]. Say that H does not contain any of K4+ ⊎ C5 , K4+ ⊎ B, 2C5 and B ⊎ C5 . Let V (D) = {x0 , x1 , x2 , x3 , x4 }, E(D) = {x0 x1 , x1 x2 , x2 x3 , x3 x4 , x4 x1 , x2 x4 } and L = a1 a2 a3 a4 a5 a1 , Set Q = [x1 , x2 , x3 , x4 ]. Since H 6⊇ 2C5 and H 6⊇ B ⊎ C5 , we z see that for each ai ∈ V (L), if x0 → (L, ai ) or x0 → (L, ai ) then e(ai , Q) ≤ 2. Thus x0 6→ L for otherwise e(D, L) ≤ 15. Hence e(x0 , L) ≤ 4. Assume e(x0 , L) = 4. Say e(x0 , a1 a2 a3 a4 ) = 4. As x0 6→ L, τ (a5 , L) = 0. Clearly, e(ai , Q) ≤ 2 for each i ∈ {2, 3, 5} since H 6⊇ 2C5 . Thus e(a1 a4 , Q) ≥ 6. Say e(a1 , Q) ≥ 3. Then [Q + a1 ] ⊇ C with C ∼ = C5 and τ (C) ≥ 3. Then a2 a4 6∈ E + for otherwise [L − a1 + x0 ] ⊇ K4 . Thus τ (L) ≤ 2. As [L − a1 + x0 ] ⊇ F1 , we see that 2 ≥ τ (L) ≥ τ (C) ≥ 3 by the optimality of {D, L}, a contradiction. Therefore e(x0 , L) ≤ 3 and so e(Q, L) ≥ 13. Set T = x2 x3 x4 x2 . We divide the proof into the following six cases.

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Case 1. N (x0 , L) = {ai , ai+1 , ai+2 } for some i ∈ {1, 2, 3, 4, 5}. Say N (x0 , L) = {a1 , a2 , a3 }. Then Q + a2 6⊇ C5 and so e(a2 , Q) ≤ 2. As x0 6→ L, z we see that τ (a2 , L) ≤ 1. If {a1 a4 , a3 a5 } ⊆ E then x0 → (L, ai ) or x0 → (L, ai ) and so e(ai , Q) ≤ 2 for each ai ∈ V (L). Consequently, e(Q, L) ≤ 10, a contradiction. Hence a1 a4 6∈ E or a3 a5 6∈ E. Thus τ (L) ≤ 3. Suppose τ (a2 , L) = 1. Say w.l.o.g. a2 a4 ∈ E. Then x0 → (L, ai ) for i ∈ {3, 5}. Thus e(ai , Q) ≤ 2 for i ∈ {3, 5}. As e(Q, L) ≥ 13, e(a1 a4 , Q) ≥ 7. Thus [Q + ar ] contains a 5-cycle with at least 4 chords, where e(ar , Q) = 4 with r ∈ {1, 4}. As [L − ar + x0 ] ⊇ F1 and by the optimality of {D, L}, we have τ (L) ≥ 4, a contradiction. Hence τ (a2 , L) = 0. Suppose a1 a3 ∈ E. Then [L − ai + x0 ] ⊇ K4+ for each i ∈ {4, 5}. As H 6⊇ K4+ ⊎ C5 , e(ai , Q) ≤ 2 for i ∈ {4, 5}. As e(Q, L) ≥ 13, e(a1 a3 , Q) ≥ 7 and e(a4 a5 , Q) ≥ 3. Say w.l.o.g. e(a5 , Q) = 2. As [Q + a5 ] 6⊇ C5 , e(a5 , x2 x4 ) = 2. As e(x1 , a1 a3 ) ≥ 1, [x1 , x0 , a1 , a2 , a3 ] ⊇ C5 . Thus e(a4 , T ) = 0 as H 6⊇ 2C5 . It follows that e(a1 a3 , Q) = 8 and a4 x1 ∈ E. Consequently, H ⊇ 2C5 , a contradiction. Hence a1 a3 6∈ E and so τ (L) ≤ 1. Since [L − ai + x0 ] ⊇ F1 for each i ∈ {4, 5}, we see that [Q + ai ] does not contain a 5-cycle with at least 2 chords for each i ∈ {4, 5} by the optimality of {D, L}. This implies that for each i ∈ {4, 5}, e(ai , Q) ≤ 2 and if e(ai , Q) = 2 then e(ai , x2 x4 ) = 2. Similar to the above, we see that H ⊇ 2C5 , a contradiction. Case 2. N (x0 , L) = {ai , ai+1 , ai+3 } for some i ∈ {1, 2, 3, 4, 5}. Say N (x0 , L) = {a1 , a2 , a4 }. Then for each i ∈ {3, 5}, x0 → (L, ai ) and so e(ai , Q) ≤ 2. Thus e(a1 a2 a4 , Q) ≥ 13−e(a3 a5 , Q) ≥ 9. Suppose that e(a3 , Q) = 2 or e(a5 , Q) = 2. Say w.l.o.g. e(a5 , Q) = 2. Then e(a5 , x2 x4 ) = 2 as [Q + a5 ] 6⊇ C5 . If a3 x3 ∈ E then [a3 , a4 , a5 , x3 , xi ] ⊇ C5 for i ∈ {2, 4} and so e(xi , a1 a2 ) = 0 for i ∈ {2, 4} since H 6⊇ 2C5 . Consequently, e(a1 a2 a4 , Q) ≤ 8, a contradiction. Hence a3 x3 6∈ E. If a3 x1 ∈ E then [x1 , x0 , a1 , a2 , a3 ] ⊇ C5 and so e(a4 , T ) = 0 as H 6⊇ 2C5 . Thus e(a1 a2 a4 , Q) = 9 and so e(a3 , Q) = 2. Consequently, [Q + a3 ] ⊇ C5 , a contradiction. Hence N (a3 , Q) ⊆ {x2 , x4 }. If e(x1 , a2 a4 ) ≥ 1 then [x1 , x0 , a2 , a3 , a4 ] ⊇ C5 and so e(a1 , T ) = 0 as H 6⊇ 2C5 . It follows that e(a3 , x2 x4 ) = 2 and e(a2 a4 , Q) = 8. Consequently, H ⊇ 2C5 , a contradiction. Hence e(x1 , a2 a4 ) = 0. Thus e(a2 a4 , T ) ≥ 5 as e(a1 a2 a4 , Q) ≥ 9. Hence [x3 , x4 , a2 , a3 , a4 ] ⊇ C5 and [x0 , x1 , x2 , a5 , a1 ] ⊇ C5 , a contradiction. Therefore e(a3 , Q) ≤ 1 and e(a5 , Q) ≤ 1. Then e(a1 a2 a4 , Q) ≥ 11. Thus e(a1 a2 , Q) ≥ 7. Say w.l.o.g. e(a1 , Q) = 4. Then [a5 , a1 , x2 , x3 , x4 ] ⊇ K4+ . As e(x1 , a2 a4 ) ≥ 1, [x1 , x0 , a2 , a3 , a4 ] ⊇ C5 and so H ⊇ K4+ ⊎ C5 , a contradiction. Case 3. N (x0 , L) = {ai , ai+1 } for some i ∈ {1, 2, 3, 4, 5}. In this case, e(Q, L) ≥ 14. Say e(x0 , a1 a2 ) = 2. Suppose x1 a4 ∈ E. Then [x1 , x0 , a1 , a5 , a4 ] ⊇ C5 . As H does not contain one of 2C5 and K4+ ⊎ C5 , we see that e(a2 a3 , T ) ≤ 2. Similarly, e(a1 a5 , T ) ≤ 2 as [x1 , x0 , a2 , a3 , a4 ] ⊇ C5 . Thus e(Q, L) ≤ 12, a contradiction. Hence x1 a4 6∈ E. Next, suppose that e(x1 , a3 a5 ) ≥ 1. Say w.l.o.g. x1 a3 ∈ E. Then [x1 , x0 , a1 , a2 , a3 ] ⊇ C5 . As H does not contain one of 2C5 ,

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B ⊎ C5 and K4+ ⊎ C5 , we have that e(a4 a5 , T ) ≤ 2 and either e(a4 , T ) = 0 or e(a5 , T ) = 0. If we also have x1 a5 ∈ E then e(a3 a4 , T ) ≤ 2 and either e(a4 , T ) = 0 or e(a3 , T ) = 0. Consequently, it follows, as e(Q, L) ≥ 14, that e(a5 , T ) = 2, e(a3 , T ) = 2, e(a4 , T ) = 0 and e(a1 a2 , Q) = 8. Then xi → (L, a1 ) for some xi ∈ V (T ) with e(xi , a2 a5 ) = 2 and so H ⊇ 2C5 , a contradiction. Hence x1 a5 6∈ E. Thus e(a1 a2 a3 , Q) ≥ 12. Then x3 → (L, a2 ) and so H ⊇ 2C5 , a contradiction. We conclude that e(x1 , a3 a4 a5 ) = 0. As e(Q, L) ≥ 14, e(x2 x4 , a1 a2 ) ≥ 1. Say w.l.o.g. e(x2 , a1 a2 ) ≥ 1. Then [x2 , x1 , x0 , a1 , a2 ] ⊇ C5 . As H 6⊇ 2C5 and by Lemma 2.1(c), e(x3 x4 , a3 a4 a5 ) ≤ 4. Thus e(a3 a4 a5 , Q) ≤ 7. Hence e(a1 a2 , Q) ≥ 7. Say w.l.o.g. e(a1 , Q) = 4. Then xi 6→ (L, a1 ) for each xi ∈ V (T ) since H 6⊇ 2C5 . This implies that I(a2 a5 , T ) = ∅ and so e(a2 a5 , Q) ≤ 4. Consequently, e(a3 a4 , T ) = 6 as e(Q, L) ≥ 14. Thus [a5 , a4 , a3 , x3 , x4 ] ⊇ K4+ and [x2 , x1 , x0 , a2 , a1 ] ⊇ C5 , a contradiction. Case 4. N (x0 , L) = {ai , ai+2 } for some i ∈ {1, 2, 3, 4, 5}. Say, N (x0 , L) = {a1 , a3 }. The e(a2 , Q) ≤ 2 as H 6⊇ 2C5 . First, suppose e(x1 , a1 a3 ) ≥ 1. Then [x1 , x0 , a1 , a2 , a3 ] ⊇ C5 and therefore e(a4 a5 , T ) ≤ 2. Thus e(a1 a3 , Q) ≥ 14 − 2 − 2 − e(x1 , a4 a5 ) ≥ 8. It follows that e(a1 a3 , Q) = 8, e(a2 , Q) = 2, e(a4 a5 , T ) = 2 and e(x1 , a4 a5 ) = 2. Consequently, H ⊇ 2C5 , a contradiction. Hence e(x1 , a1 a3 ) = 0. Next, suppose e(x1 , a4 a5 ) ≥ 1. Say w.l.o.g. x1 a4 ∈ E. Then [x1 , x0 , a1 , a5 , a4 ] ⊇ C5 and so e(a2 a3 , T ) ≤ 2. Thus e(a1 a5 a4 , Q) ≥ 14 − 3 = 11. It follows that e(a4 a5 , Q) = 8, e(a1 , T ) = 3, x1 a2 ∈ E and e(a2 a3 , T ) = 2. Then [D − x1 + a1 ] ⊇ K4+ and [L − a1 + x1 ] ⊇ C5 , a contradiction. Hence e(x1 , a4 a5 ) = 0. As e(Q, L) ≥ 14, it follows that e(T, L − a2 ) = 12 and e(a2 , Q) = 2. Then we readily see that H ⊇ 2C5 , a contradiction. Case 5. e(x0 , L) = 1. Then e(Q, L) ≥ 15. Say x0 a1 ∈ E. First, suppose e(x1 , a3 a4 ) ≥ 1. Say w.l.o.g. x1 a3 ∈ E. Then [x1 , x0 , a1 , a2 , a3 ] ⊇ C5 . Thus e(a4 a5 , T ) ≤ 2 and so e(a4 a5 , Q) ≤ 4. If we also have x1 a4 ∈ E then e(a2 a3 , T ) ≤ 2 as [x1 , x0 , a1 , a5 , a4 ] ⊇ C5 . But then we obtain e(Q, L) ≤ 12, a contradiction. Hence x1 a4 6∈ E. As e(Q, L) ≥ 15, it follows that e(a1 a2 a3 , Q) = 12, e(a4 a5 , T ) = 2 and x1 a5 ∈ E. Then [a4 , a5 , x1 , x0 , a1 ] ⊇ F1 and [T, a2 , a3 ] ⊇ K5 . By the optimality of {D, L}, [L] ∼ = K5 and so H ⊇ 2C5 , a contradiction. Hence e(x1 , a3 a4 ) = 0. Then e(a2 a5 , Q) ≥ 15 − e(a1 a3 a4 , Q) ≥ 15 − 10 = 5. Thus e(x2 x4 , a2 a5 ) ≥ 1. Say w.l.o.g. x2 a5 ∈ E. Then [x0 , x1 , x2 , a5 , a1 ] ⊇ C5 . As H 6⊇ 2C5 , e(a2 a4 , x3 x4 ) ≤ 2. Clearly, e(a2 a3 a4 , x1 x2 ) ≤ 4. Then e(a1 a5 , Q) ≥ 15 − 6 − e(a3 , x3 x4 ) ≥ 7 and so e(a1 , T ) ≥ 2. Suppose that a1 x3 ∈ E. Then xi 6→ (L, a1 ) for all xi ∈ V (T ) for otherwise H ⊇ 2C5 . This implies that I(a2 a5 , T ) = ∅. As x2 a5 ∈, x2 a2 6∈ E and so e(a2 a3 a4 , x1 x2 ) ≤ 3. As e(Q, L) ≥ 15, it follows that e(a1 a5 , Q) = 8, e(a2 a3 a4 , x3 x4 ) = 4 and so e(x3 x4 , a3 a4 ) = 4. Thus [a2 , a3 , a4 , x3 , x4 ] ⊇ K4+ and so H ⊇ K4+ ⊎ C5 , a contradiction. Hence a1 x3 6∈ E. Thus e(a1 a5 , Q) = 7. It follows that e(a1 , Q − x3 ) = 3, e(a5 , Q) = 4, e(a2 a4 , x3 x4 ) = 2, e(a3 , x3 x4 ) = 2, e(x2 , a3 a4 ) = 2 and e(a2 , x1 x2 ) = 2. Then

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[x2 , x1 , x0 , a1 , a2 ] ⊇ C5 and [a5 , a4 , a3 , x3 , x4 ] ⊇ C5 , a contradiction. Case 6. e(x0 , L) = 0. As H 6⊇ K4+ ⊎ C5 , we see that for each ai ∈ V (L), if e(ai , Q − x3 ) = 3 then x3 6→ (L, ai ). Since e(ai , Q) = 4 for some ai ∈ V (L) as e(Q, L) ≥ 16, it follows that x3 6→ L and so e(x3 , L) ≤ 4. First, suppose e(x3 , L) = 4. Say e(x3 , L − a5 ) = 4. Then e(ai , Q − x3 ) ≤ 2 for each i ∈ {2, 3, 5}. As e(Q, L) ≥ 16, it follows that e(ai , Q−x3 ) = 2 for i ∈ {2, 3, 5} and e(a1 a4 , Q − x3 ) = 6. If x1 a5 ∈ E, then e(a5 , x1 x2 ) = 2 or e(a5 , x1 x4 ) = 2. Say w.l.o.g. e(a5 , x1 x2 ) = 2. Then [x0 , x1 , x2 , a1 , a5 ] ⊇ K4+ and [x3 , x4 , a2 , a3 , a4 ] ⊇ C5 , a contradiction. Hence e(a5 , x2 x4 ) = 2. Then [D − x3 + a5 ] ⊇ F1 . By the optimality of {D, L}, τ (L) ≥ τ (x3 a1 a2 a3 a4 x3 ). This implies that τ (a5 , L) = 2 and so x3 → (L, a1 ), a contradiction. Next, suppose that e(x3 , L) = 3 and N (x3 , L) = {ai , ai+1 , ai+3 } for some i ∈ {1, 2, 3, 4, 5}. Say N (x3 , L) = {a1 , a2 , a4 }. Then e(a3 , Q − x3 ) ≤ 2 and e(a5 , Q − x3 ) ≤ 2. As e(Q, L) ≥ 16, it follows that e(a1 a2 a4 , Q − x3 ) = 9, e(a3 , Q − x3 ) = 2 and e(a5 , Q − x3 ) = 2. If e(x1 , a3 a5 ) ≥ 1, then we may assume w.l.o.g. that e(a3 , x1 x2 ) = 2. Consequently, [x0 , x1 , x2 , a2 , a3 ] ⊇ K4+ and [x3 , x4 , a1 , a5 , a4 ] ⊇ C5 , a contradiction. Hence e(a3 a5 , x2 x4 ) = 4. Clearly, [x0 , x1 , x2 , a2 , a3 ] ⊇ F1 and τ (x4 x3 a1 a5 a4 x4 ) ≥ 3. Thus τ (L) ≥ 3 by the optimality of {D, L}. As x3 6→ (L, a1 ), a3 a5 6∈ E. Thus a1 a4 ∈ E or a2 a4 ∈ E. Say w.l.o.g. a1 a4 ∈ E. Then τ (x4 x3 a1 a5 a4 x4 ) = 4. Thus τ (L) = 4 and so the lemma holds. Next, suppose that N (x3 , L) = {ai , ai+1 , ai+2 } for some i ∈ {1, 2, 3, 4, 5}. Say N (x3 , L) = {a1 , a2 , a3 }. Then e(a2 , Q − x3 ) ≤ 2. As e(D, L) ≥ 16, either e(a1 a5 , Q−x3 ) = 6 or e(a3 a4 , Q−x3 ) = 6. Say w.l.o.g. e(a1 a5 , Q−x3 ) = 6. Then [x0 , x1 , xi , a1 , a5 ] ⊇ K4+ and so [x3 , xj , a2 , a3 , a4 ] 6⊇ C5 for each {i, j} = {2, 4}. This implies that e(a4 , x2 x4 ) = 0 and so e(D, L) ≤ 15, a contradiction. Next, suppose that e(x3 , L) = 2 and N (x3 , L) = {ai , ai+2 } for some i ∈ {1, 2, 3, 4, 5}. Say N (x3 , L) = {a1 , a3 }. Then e(a2 , Q − x3 ) ≤ 2. As e(Q, L) ≥ 16, it follows that e(L − a2 , Q − x3 ) = 12 and e(a2 , Q − x3 ) = 2. Then [x0 , x1 , x2 , a4 , a5 ] ⊇ K4+ and [x3 , x4 , a1 , a2 , a3 ] ⊇ C5 , a contradiction. Next, suppose that e(x3 , L) = 2 and N (x3 , L) = {ai , ai+1 } for some i ∈ {1, 2, 3, 4, 5}. Say N (x3 , L) = {a1 , a2 }. As e(Q, L) ≥ 16, either e(a1 a5 , Q−x3 ) = 6 or e(a2 a3 , Q−x3 ) = 6. Say w.l.o.g. e(a1 a5 , Q−x3 ) = 6. Then [x0 , x1 , xi , a1 , a5 ] ⊇ K4+ and so [xj , x3 , a2 , a3 , a4 ] 6⊇ C5 for each {i, j} = {2, 4}. This implies that e(a4 , x2 x4 ) = 0. Consequently, e(Q, L) ≤ 15, a contradiction. Finally, we have e(x3 , L) = 1. Then e(L, Q − x3 ) = 15, clearly, H ⊇ K4+ ⊎ C5 , a contradiction. Lemma 2.10. Let D, L1 and L2 be disjoint subgraphs of G with D ∼ = F1 and L1 ∼ = L2 ∼ = C5 . Suppose that L1 = a1 a2 a3 a4 a5 a1 , V (D) = {x0 , x1 , x2 , x3 , x4 } and E(D) = {x0 x1 , x1 x2 , x2 x3 , x3 x4 , x4 x1 , x2 x4 } such that

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e(x0 , L1 ) = 0, e(a1 a2 a4 , D −x0 ) = 12, N (a3 , D) = N (a5 , D) = {x2 , x4 }, τ (L1 ) = 4 and a3 a5 6∈ E. Suppose that {D, L1 , L2 } is optimal and e(x0 x3 a3 a5 , L2 ) ≥ 13. Then [D, L1 , L2 ] contains either K4+ ⊎ 2C5 or 3C5 . Proof. Let G1 = [D, L1 ], G2 = [D, L1 , L2 ] and R = {x0 , x3 , a3 , a5 }. On the contrary, suppose that G2 does not contain any of K4+ ⊎ 2C5 and 3C5 . It is easy to see that for any permutation f of {x3 , a3 , a5 }, we can extend f to be an automorphism of G1 such that any vertex in G1 − {x3 , a3 , a5 } is fixed under f . Thus x3 , a3 and a5 are in the symmetric position in the following argument. It is easy to check that if u → (L2 ; R − {u}) for some u ∈ R, then G2 ⊇ K4+ ⊎ 2C5 or G2 ⊇ 3C5 . Thus u 6→ (L2 ; R − {u}) for each u ∈ R. By Lemma 2.1(d), there exist two labellings R = {y1 , y2 , y3 , y4 } and L2 = b1 b2 b3 b4 b5 b1 such that e(y1 y2 , b1 b2 b3 b4 ) = 8, e(y3 , b1 b5 b4 ) = 3 and e(y4 , b1 b4 ) = 2. If x0 ∈ {y1 , y2 }, we may assume w.l.o.g. that {x0 , x3 } = {y1 , y2 }. Then [G1 − x0 + b5 ] ⊇ F1 ⊎ K5− . na By the optimality of {D, L1 , L2 }, x0 → (L2 , b5 ). This implies that τ (b5 , L2 ) = 2. Thus x0 → (L2 , b1 ; R − {x0 }), a contradiction. Hence x0 6∈ {y1 , y2 }. W.l.o.g., say {a3 , a5 } = {y1 , y2 }. Then [a3 , a4 , a5 , b2 , b3 ] ⊇ C5 , [x0 , x3 , b1 , b5 , b4 ] ⊇ C5 and [x2 , x1 , x4 , a1 , a2 ] ⊇ C5 , a contradiction.

3.

Proof of Theorem 1

Let G be a graph of order 5k with minimum degree at least 3k. Suppose, for a contradiction, that G 6⊇ kC5 . We may assume that G is maximal, i.e., G + xy ⊇ kC5 for each pair of non-adjacent vertices x and y of G. Thus G ⊇ P5 ⊎ (k − 1)C5 . Our proof will follow from the following three lemmas. Lemma 3.1. For each s ∈ {1, 2, . . . , k}, G 6⊇ sB ⊎ (k − s)C5 . Proof. On the contrary, suppose that G ⊇ sB ⊎ (k − s) C5 for some s ∈ {1, 2, . . . , k}. Let s be the minimum number in {1, 2, . . . , k} such that G ⊇ sB ⊎(k −s)C5 . Say G ⊇ sB ⊎(k −s)C5 = {B1 , . . . , Bs , L1 , . . . , Lk−s } with Bi ∼ =B for i ∈ {1, 2, . . . , s}. Let R be the set of the four vertices of B1 whose degrees in B1 are 2. By Lemma 2.2, Lemma 2.8 and the minimality of s, we see that e(R, Bi ) ≤ 12 and e(R, Lj ) ≤ 12 for all i ∈ {2, 3, . . . , s} and j ∈ {1, 2, . . . , k − s}. Therefore e(R, G) ≤ 12(k − 1) + 8 = 12k − 4. As the minimum degree of G is 3k, we obtain 12k − 4 ≥ e(R, G) ≥ 12k, a contradiction. Lemma 3.2. There exists a sequence (D, L1 , L2 , . . . , Lk−1 ) of disjoint subgraphs of G such that D ∼ = K4+ and Li ∼ = C5 for all i ∈ {1, 2, . . . , k − 1}. Proof. First, we claim that G ⊇ F ⊎ (k − 1) C5 . We choose a sequence (P,L1 ,L2 , . . . , Lk−1 ) of disjoint subgraphs of G such that P ∼ = C5 for = P5 and Li ∼

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Pk−1 all i ∈ {1, 2, . . . , k − 1} with i=1 τ (Li ) as large as possible. As G 6⊇ kC5 and by Lemma 2.1(c), e(P, P ) ≤ 14 and so e(P, G − V (P )) ≥ 15k − 14 = 15(k − 1) + 1. Thus e(P, Li ) ≥ 16 for some i ∈ {1, 2, . . . , k − 1}. By Lemma 2.3, [P, Li ] ⊇ F ⊎ C5 and so G ⊇ F ⊎ (k − 1)C5 . Next, we claim that G ⊇ F1 ⊎ (k − 1)C5 . Assume for the moment that G ⊇ F2 ⊎ (k − 1)C5 = {D, L1 , L2 , . . . , Lk−1 } with D ∼ = F2 . Let R be the three vertices of D with degree 2 in D. Then e(R, G − V (D)) ≥ 9k − 6 = 9(k − 1) + 3. Thus e(R, Li ) ≥ 10 for some i ∈ {1, 2, . . . , k − 1}. By Lemma 2.4, [D, Li ] ⊇ F1 ⊎ C5 and so G ⊇ F1 ⊎ (k − 1)C5 . Hence we may assume that G 6⊇ F2 ⊎ (k − 1)C5 . Then we choose a sequence (D, L1 , L2 , . . . , Lk−1 ) of disjoint subgraphs of G such that D ∼ = C5 for all i ∈ {1, 2, . . . , k − 1} with = F and Li ∼ Pk−1 τ (L ) as large as possible. Then e(D, Li ) ≥ 16 for some i ∈ {1, 2, . . . , k −1}. i i=1 By Lemma 2.5 and Lemma 3.1, we may assume that there exist two labellings D = x0 x1 x2 x3 x4 x1 and L1 = a1 a2 a3 a4 a5 a1 such that e(x0 , L1 ) = 0, e(x1 x3 , L1 ) = 10, N (x2 , L1 ) = N (x4 , L1 ) = {a1 , a2 , a4 }, τ (L1 ) = 4 and a3 a5 6∈ E. Then e(x0 x2 a3 a5 , G−V (D∪L1 )) ≥ 12k −17 = 12(k −2)+7. Thus e(x0 x2 a3 a5 , Li ) ≥ 13 for some i ∈ {2, 3, . . . , k − 1}. By Lemma 2.6, we obtain [D, L1 , Li ] ⊇ F1 ⊎ 2C5 and so G ⊇ F1 ⊎ (k − 1)C5 . Suppose that G ⊇ K4+ ⊎ B ⊎ (k − 2)C5 = {D, B1 , L1 , L2 , . . . , Lk−2 } with D ∼ = B. Let R be the four vertices of B1 with degree 2 in B1 . = K4+ and B1 ∼ Then either e(R, D) ≥ 13 or e(R, Li ) ≥ 13 for some i ∈ {1, 2, . . . , k − 2}. By Lemma 2.2, Lemma 2.7 and Lemma 3.1, we see that G ⊇ K4+ ⊎ (k − 1)C5 . Hence we may suppose that G 6⊇ K4+ ⊎ B ⊎ (k − 2)C5 . We now choose an optimal sequence (D, L1 , L2 , . . . , Lk−1 ) of disjoint subgraphs of G with D ∼ = C5 for all i ∈ {1, 2, . . . , k−1}. Then e(D, Li ) ≥ = F1 and Li ∼ 16 for some i ∈ {1, 2, . . . , k − 1}. Say w.l.o.g. e(D, L1 ) ≥ 16. By Lemma 2.9 and Lemma 3.1, we may assume that there exist two labellings L1 = a1 a2 a3 a4 a5 a1 and V (D) = {x0 , x1 , x2 , x3 , x4 } with E(D) = {x0 x1 , x1 x2 , x2 x3 , x3 x4 , x4 x1 , x2 x4 } such that e(x0 , L1 ) = 0, e(a1 a2 a4 , D − x0 ) = 12, N (a3 , L1 ) = N (a5 , L1 ) = {x2 , x4 }, τ (L1 ) = 4 and a3 a5 6∈ E. Let R = {x0 , x3 , a3 , a5 } and G1 = [D, L1 ]. Then e(R, G1 ) ≤ 16 and so e(R, G − V (G1 )) ≥ 12k − 16 = 12(k − 2) + 8. This implies that e(R, Li ) ≥ 13 for some i ∈ {2, 3, . . . , k − 1}. Say w.l.o.g. e(R, L2 ) ≥ 13. By Lemma 2.10, it follows that [G1 , L2 ] ⊇ K4+ ⊎ 2C5 and so G ⊇ K4+ ⊎ (k − 1)C5 . Let σ = (D, L1 , . . . , Lk−1 ) be an optimal sequence of disjoint subgraphs in G with D∼ = C5 for all i ∈ {1, 2, . . . , k − 1}. Say V (D) = {x0 , x1 , x2 , x3 , x4 } = K4+ and Li ∼ with N (x0 , D) = {x1 }. Let Q = D − x0 and T = Q − x1 . Then Q ∼ = K4 and C . T ∼ = 3 Lemma 3.3. For each t ∈ {1, 2, . . . , k − 1}, the following statements hold: (a) If e(x0 , Lt ) = 5, then e(Q, Lt ) ≤ 5.

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(b) If e(x0 , Lt ) = 4, then e(Q, Lt ) ≤ 9. (c) If e(x0 , Lt ) = r, then e(Q, Lt ) ≤ 18 − 2r for r ∈ {1, 3} and if e(x0 , Lt ) = 2, then e(Q, Lt ) ≤ 15. Proof. For convenience, we may assume Lt = L1 = a1 a2 a3 a4 a5 a1 . Let G1 = [D, L1 ]. As G1 6⊇ 2C5 , we see that if x0 → L1 , then e(ai , Q) ≤ 1 for all ai ∈ V (L1 ) and so the lemma holds. Hence we may assume that x0 6→ L1 and so e(x0 , L1 ) ≤ 4. To prove (b), say w.l.o.g. e(x0 , L1 − a5 ) = 4. On the contrary, suppose e(Q, L1 ) ≥ 10. It is easy to see that τ (a5 , L1 ) = 0 for otherwise x0 → L1 and so G1 ⊇ 2C5 . As x0 → (L1 , ai ) for i ∈ {2, 3, 5}, e(ai , Q) ≤ 1 for i ∈ {2, 3, 5}. If e(a5 , Q) = 1 then [Q + a5 ] ∼ = K4+ and τ (x0 a1 a2 a3 a4 x0 ) > τ (L1 ), contradicting the optimality of σ. Hence e(a5 , Q) = 0. It follows that e(a2 , Q) = e(a3 , Q) = 1 and e(a1 a4 , Q) = 8. Clearly, τ (x0 a3 a4 a5 a1 x0 ) ≥ τ (L1 ) with equality only if a2 a4 ∈ E. As [Q + a2 ] ⊇ K4+ and by the optimality of σ, we obtain a2 a4 ∈ E. Thus [a5 , a4 , a3 , a2 , x0 ] ⊇ K4+ and [Q + a1 ] ∼ = K5 . By the optimality of σ, we K , a contradiction. obtain [L1 ] ∼ = 5 To prove (c), we suppose, for a contradiction, that either e(x0 , L1 ) = r and e(Q, L1 ) ≥ 19 − 2r for some r ∈ {1, 3} or e(x0 , L1 ) = 2 and e(Q, L1 ) ≥ 16. We divide the proof into the following three cases. Case 1. e(x0 , L1 ) = 3 and e(Q, L1 ) ≥ 13. First, suppose that N (x0 , L1 ) = {ai , ai+1 , ai+3 } for some i ∈ {1, 2, 3, 4, 5}. Say w.l.o.g. N (x0 , L1 ) = {a1 , a2 , a4 }. As x0 6→ L1 , a3 a5 6∈ E. Clearly, x0 → (L1 , a3 ) and x0 → (L1 , a5 ). Thus e(a3 , Q) ≤ 1 and e(a5 , Q) ≤ 1. It follows that e(a1 a2 a4 , Q) ≥ 11, e(x1 , a1 a4 ) ≥ 1 and e(x1 , a2 a4 ) ≥ 1. Thus [x0 , x1 , a1 , a5 , a4 ] ⊇ C5 and [x0 , x1 , a2 , a3 , a4 ] ⊇ C5 . As e(ai , T ) ≥ 2 for i ∈ {1, 2}, it is easy to see that e(a3 a5 , T ) = 0, i.e., N (a3 a5 , Q) ⊆ {x1 }, for otherwise G1 ⊇ 2C5 . Let R = {x0 , x3 , a3 , a5 }. Then e(R, G1 ) ≤ 18 and so e(R, G − V (G1 )) ≥ 12k − 18 = 12(k − 2) + 6. Then e(R, Li ) ≥ 13 for some i ∈ {2, 3, . . . , k − 1}. Say w.l.o.g. e(R, L2 ) ≥ 13. Let G2 = [G1 , L2 ]. Then G2 6⊇ 3C5 . Since e(Q, L1 ) ≥ 13 and N (a3 a5 , Q) ⊆ {x1 }, it is easy to check that if u → (L2 ; R − {u}) for some u ∈ R, then G2 ⊇ 3C5 . Hence u 6→ (L2 ; R − {u}) for all u ∈ R. By Lemma 2.1(d), there exist two labellings L2 = b1 b2 b3 b4 b5 b1 and R = {y1 , y2 , y3 , y4 } such that e(y1 y2 , L2 −b5 ) = 8, e(y3 , b1 b5 b4 ) = 3 and e(y4 , b1 b4 ) = 2. If {y1 , y2 } = {x0 , x3 }, let {s, t} = {1, 2} with as ∈ I(x0 x3 , L1 ) and then we see that [x0 , as , x3 , b2 , b3 ] ⊇ C5 , [a3 , a5 , b1 , b5 , b4 ] ⊇ C5 and [Q − x3 + a4 + at ] ⊇ C5 , a contradiction. If {y1 , y2 } = {x0 , ai } for some i ∈ {3, 5}, we may assume w.l.o.g. that {y1 , y2 } = {x0 , a5 } and then we see that [x0 , a1 , a5 , b2 , b3 ] ⊇ C5 , [a3 , x3 , b1 , b5 , b4 ] ⊇ C5 and [a2 , a4 , x1 , x2 , x4 ] ⊇ C5 , a contradiction. If {y1 , y2 } = {x3 , ai } for some i ∈ {3, 5}, we may assume w.l.o.g. that {y1 , y2 } = {x3 , a5 } and let {s, t} = {1, 4} be such that x3 as ∈ E. Then we see that {x3 , as , a5 , b2 , b3 ] ⊇

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C5 , [x0 , a3 , b1 , b5 , b4 ] ⊇ C5 and [x1 , x2 , x4 , a2 , at ] ⊇ C5 , a contradiction. Hence {y1 , y2 } = {a3 , a5 }. Then [a3 , a4 , a5 , b2 , b3 ] ⊇ C5 , [x0 , x3 , b1 , b5 , b4 ] ⊇ C5 and [x1 , x2 , x4 , a1 , a2 ] ⊇ C5 , a contradiction. Next, suppose that N (x0 , L1 ) = {ai , ai+1 , ai+2 } for some i ∈ {1, 2, 3, 4, 5}. Say w.l.o.g. N (x0 , L1 ) = {a1 , a2 , a3 }. Then e(a2 , Q) ≤ 1 as G1 6⊇ 2C5 and so e(Q, L1 − a2 ) ≥ 12. First, assume e(x1 , a4 a5 ) ≥ 1. Say w.l.o.g. x1 a5 ∈ E. Then [x0 , x1 , a5 , a1 , a2 ] ⊇ C5 . Then e(a3 a4 , T ) ≤ 3 as G1 6⊇ 2C5 . If we also have x1 a4 ∈ E, then similarly, e(a1 a5 , T ) ≤ 3 and so e(Q, L1 − a2 ) ≤ 11, a contradiction. Hence x1 a4 6∈ E. As e(Q, L1 ) ≥ 13, it follows that e(a1 a5 , Q) = 8, e(a3 a4 , T ) = 3, x1 a3 ∈ E and e(a2 , Q) = 1. Clearly, [T + a4 + a5 ] 6⊇ C5 as G1 6⊇ 2C5 . This implies that e(a4 , T ) = 0 and so e(a3 , Q) = 4. Obviously, G1 ⊇ 2C5 , a contradiction. Hence e(x1 , a4 a5 ) = 0. Next, assume e(x1 , a1 a3 ) ≥ 1. Then [x0 , x1 , a1 , a2 , a3 ] ⊇ C5 and so e(a4 a5 , T ) ≤ 3. It follows that e(Q, L1 − a2 ) ≤ 12, a contradiction. Hence e(x1 , L1 − a2 ) = 0. Thus e(T, L1 − a2 ) = 12. Obviously, G1 ⊇ 2C5 , a contradiction. Case 2. e(x0 , L1 ) = 2 and e(Q, L1 ) ≥ 16. First, suppose that N (x0 , L1 ) = {ai , ai+2 } for some i ∈ {1, 2, 3, 4, 5}. Say, N (x0 , L1 ) = {a1 , a3 }. Then e(a2 , Q) ≤ 1 and e(Q, L1 − a2 ) ≥ 15. Thus e(x1 , a1 a3 ) ≥ 1. Then [x0 , x1 , a1 , a2 , a3 ] ⊇ C5 and so e(a4 a5 , T ) ≤ 3. Thus e(Q, L1 − a2 ) ≤ 13, a contradiction. Therefore we may assume w.l.o.g. that N (x0 , L1 ) = {a1 , a2 }. First, assume x1 a4 ∈ E. Then [x0 , x1 , a4 , a5 , a1 ] ⊇ C5 and [x0 , x1 , a4 , a3 , a2 ] ⊇ C5 . As G1 6⊇ 2C5 , e(a2 a3 , T ) ≤ 3 and e(a1 a5 , T ) ≤ 3. Thus e(Q, L1 ) ≤ 14, a contradiction. Hence x1 a4 6∈ E. Next, assume e(x1 , a3 a5 ) ≥ 1. Say w.l.o.g. x1 a5 ∈ E. Then [x0 , x1 , a5 , a1 , a2 ] ⊇ C5 and so e(a3 a4 , T ) ≤ 3. As e(Q, L1 ) ≥ 16, it follows that e(a5 a1 a2 , Q) = 12, e(a3 a4 , T ) = 3 and x1 a3 ∈ E. Thus e(x3 , a2 a5 ) = 2 and so G1 ⊇ 2C5 , a contradiction. Hence e(x1 , a3 a4 a5 ) = 0. Thus e(T, L1 ) ≥ 14. This implies that e(xi , a2 a5 ) = 2 and a1 xj ∈ E for some {i, j} ⊆ {2, 3, 4} with i 6= j. Consequently, H ⊇ 2C5 , a contradiction. Case 3. e(x0 , L1 ) = 1 and e(Q, L1 ) ≥ 17. Say w.l.o.g. x0 a1 ∈ E. Suppose e(x1 , a3 a4 ) ≥ 1. Say x1 a3 ∈ E. Then [x1 , x0 , a1 , a2 , a3 ] ⊇ C5 and so e(a4 a5 , T ) ≤ 3 as G1 6⊇ 2C5 . As e(Q, L1 ) ≥ 17, it follows that e(a1 a2 a3 , Q) = 12, e(a4 a5 , T ) = 3 and e(x1 , a4 a5 ) = 2. Then [x0 , x1 , a4 , a5 , a1 ] ⊇ C5 and [T, a2 , a3 ] ⊇ C5 , a contradiction. Hence e(x1 , a3 a4 ) = 0. As e(Q, L1 ) ≥ 17, e(T, L1 ) ≥ 14. This implies that e(xi , a2 a5 ) = 2 and a1 xj ∈ E for some {i, j} ⊆ {2, 3, 4} with i 6= j. Consequently, H ⊇ 2C5 , a contradiction. We are now in the position to complete the proof of Theorem 1. Let Ar = {Lt |e(x0 , Lt ) = r, 1 ≤ t ≤ k − 1} for each 0 ≤ r ≤ 5. Set pr = |Ar | for each 0 ≤ r ≤ 5. Clearly, p0 + p1 + p2 + p3 + p4 + p5 = k − 1. By Lemma 3.3, we obtain

Disjoint 5-cycles in a Graph

e(x0 , G) = e(x0 , D) +

241

5 X X

e(x0 , Lt )

r=0 Lt ∈Ar

(2)

= 1 + p1 + 2p2 + 3p3 + 4p4 + 5p5 ; 5 X X e(D, G) = e(D, D) + e(D, Lt ) r=0 Lt ∈Ar

≤ 14 + 20p0 + 17p1 + 17p2 + 15p3 + 13p4 + 10p5 .

(3) Then we obtain

e(x0 , G) + e(D, G) ≤ 15 + 20p0 + 18p1 + 19p2 + 18p3 + 17p4 + 15p5 = 18k + 2p0 + p2 − p4 − 3p5 − 3.

(4) As 3

P5

r=0 pr

= 3k − 3 and e(x0 , G) ≥ 3k, we obtain, by using (2), the following 1 + p1 + 2p2 + 3p3 + 4p4 + 5p5

(5)

≥ 3 + 3p0 + 3p1 + 3p2 + 3p3 + 3p4 + 3p5 .

This implies that 3p0 + 2p1 + p2 − p4 − 2p5 + 2 ≤ 0. Thus 2p0 + p2 − p4 − 3p5 ≤ −2. Together with (4), we obtain e(x0 , G) + e(D, G) ≤ 18k − 5. But by the degree condition on G, we have e(x0 , G) + e(D, G) ≥ 3k + 15k = 18k, a contradiction. This proves Theorem 1. References [1] S. Abbasi, PhD Thesis (Rutgers University 1998). [2] B. Bollob´as, Extremal Graph Theory (Academic Press, London, 1978). [3] K. Corr´adi and A. Hajnal, On the maximal number of independent circuits in a graph, Acta Math. Acad. Sci. Hungar. 14 (1963) 423–439. doi:10.1007/BF01895727 [4] M.H. El-Zahar, On circuits in graphs, Discrete Math. 50 (1984) 227–230. doi:10.1016/0012-365X(84)90050-5 [5] P. Erd˝ os, Some recent combinatorial problems, Technical Report, University of Bielefeld, Nov. 1990. [6] B. Randerath, I. Schiermeyer and H. Wang, On quadrilaterals in a graph, Discrete Math. 203 (1999) 229–237. doi:10.1016/S0012-365X(99)00053-9 [7] H. Wang, On quadrilaterals in a graph, Discrete Math. 288 (2004) 149–166. doi:10.1016/j.disc.2004.02.020

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[8] H. Wang, Proof of the Erd˝ os-Faudree conjecture on quadrilaterals, Graphs and Combin. 26 (2010) 833–877. doi:10.1007/s00373-010-0948-3 Received 26 October 2010 Revised 17 April 2011 Accepted 17 April 2011